Chapter 6
Work and Energy
Work (W): Work is defined as the product of parallel force and distance. W = F_{ll} x . F_{ll} denotes the component of force F that is parallel to displacement, x.
The SI unit for work is Nm called " Joule", and the American unit is " lbft ."
If the direction of F is parallel to that of x, the work done is simply W = F x as in the following example.
Example 1: Find the work done by force F = 25N in pushing the block a distance of 14m.
If an object placed on a horizontal surface is pushed or pulled by force F as shown below, the component of F that is parallel to x does useful work. The component perpendicular to x does no work.
Example 2: A block is pulled a distance of x = 24m from A to B as shown via force F = 45N that makes a 30.0° angle with the horizontal surface. Calculate the work done by F.
Solution: For θ = 30.0°, and F = 45N, F_{ll} = 45N cos(30.0 ) = 39N.
W = F_{ll} x ; W = (39N)(24m) = 940 Nm or, W = 940 J.
Example 3: In the figure shown, find (a) the magnitude of F such that the block slides at constant speed to the right. (b) Find the work done by this force if the displacement is 38 m.
Example 4: In Example 3, find the work done by the frictional force, F_{k} ,within the same distance.
Solution: F_{k} acts opposite to x; therefore, it does negative work. In other words, it consumes energy.
W_{friction} = F_{k} x ; W = (45N )( 38m ) = 1700 J.
Energy: Energy is defined as the ability to do work. Energy and work are expressed in same units. Typical units are, Joule and lbft, as were mentioned above. Other units are: Calorie and kilocalorie (that you will learn in the chapter on heat energy), and electronvolt (that will be discussed under electric energy). The abbreviations for these units are: cal, kcal, and eV.
Types of Energy: Energy (the ability of doing work) exists in different forms such as mechanical, electric, nuclear, light, chemical, etc... . In this chapter, mechanical energy will be discussed only. Heat energy is also a topic of Physics I that will be discussed in Chapter 12.
Mechanical Energy: Kinetic energy, Gravitational potential energy, and elastic (spring) potential energy are forms of mechanical energy and will be discussed under this topic.
I ) Kinetic Energy (K.E.): Kinetic energy is the energy that a mass (an object) has because of its motion. As long as an object moves or has some speed, it has kinetic energy. K.E. is proportional to mass (M) and proportional to the square of velocity (v^{2}). Its formula is: (without proof in this course)
K.E. = (1/2)Mv^{2}
Example 5: Calculate the K.E. of a 2000.kg car that is moving (a) at 10.0m/s, (b) 20.0 m/s, and (c) 30.0 m/s.
Solution: (a) (K.E.)_{1} = (1/2)(2000.kg)(10.0m/s)^{2} = 100,000 (kg m/s^{2})m = 100,000 J (3 sig. fig.)
(b) (K.E.)_{2} = (1/2)(2000.kg)(20.0m/s)^{2} = 400,000 (kg m/s^{2})m = 400,000 J (3 sig. fig.)
(c) (K.E.)_{3} = (1/2)(2000.kg)(30.0m/s)^{2} = 900,000 (kg m/s^{2})m = 900,000 J (3 sig. fig.)
As can be seen, the car's kinetic energy varies with the square of velocity (v^{2}).
II ) Gravitational Potential Energy (P.E.): This energy is the type that an object has due to its elevation with respect to a reference level. This energy is significant when objects are in the vicinity of planets, stars and other heavenly masses. Gravitational P.E. is proportional to the acceleration of gravity (g) of the planet or star, the mass of the object, M, and its elevation (h) from a reference level. Its formula is therefore,
P.E. = Mgh.
Example 6: A rock of mass 6.1 kg is raised to the top of a building 13 m high. Calculate its gravitational P.E. with respect to the ground.
Solution: Using P.E. = Mgh, yields: P.E. = (6.1 kg)(9.8 m/s^{2})(13 m) = 780 Nm = 780J.
Example 7: In the figure shown, a 95kg crate is pushed up on an incline that is practically frictionless. The incline is 8.0m long and makes a 31° angle with the horizontal floor. Calculate the P.E. of the crate with respect to the floor when it reaches the top of the incline.
Example 8: A swimming pool is on the top of a hill at an average elevation of 150m from a certain ground level. The swimming pool has dimensions: 15m X 25m X 2.4m and is full. The mass density of water is 1.000 ton/m^{3}. Determine (a) the P.E. of the water in the pool with respect to that level. If the pool is allowed to empty to the ground level, (b) how much energy is at most available for use? If a generator uses this energy for electricity production, and the overall efficiency is 62%, (c) how much electric energy will become available?
Solution: (a) The pool's volume is V = w l h ; V = 900m^{3} ; ( " r " is pronounced "rho" ).
Since density r = M / V; therefore, M = rV ; M = (1000 kg/m^{3})(900 kg) = 900,000kg.
P.E. = Mgh ; P.E. = ( 900,000 kg )( 9.8 m/s^{2} )( 150m ) = 1.3 X 10^{9} J.
(b) 1.3 X 10^{9} J
(c) 0.62 (1.3 X 10^{9} J) = 8.2 X 10^{8} J.
Example 9: A 750kg car is traveling at a velocity of 72 km/h eastward and on a level road. Determine (a) its initial kinetic energy (K.E.)_{i} . If this car is slowed down to a speed of 36 km/h, calculate (b) its final (K.E.)_{f }. (c) How much is the change in its kinetic energy and where does this energy go?
Solution: (a) (K.E.)_{i} = (1/2)Mv_{i}^{2} ; (K.E.)_{i} = (1/2)( 750 kg)( 20. m/s)^{2} = 150,000 J.
(b) (K.E.)_{f } = (1/2)Mv_{f}^{2} ; (K.E.)_{f } = (1/2)( 750 kg)( 10. m/s)^{2} = 38,000 J.
(c) Δ(K.E.) = (K.E.)_{f }  (K.E.)_{i} ; Δ(K.E.) = 38,000 J  150,000 J = 112,000 J.
This energy is consumed by force of friction (brakes force, for example) that acts opposite to the direction of motion. In fact, it is the work done by force of friction. Most of this energy converts to heat and warms up the brakes. Friction does negative work.
Example 10: In Example 8, the work done by force of friction is 112,000J. Use the work formula to calculate the force applied by friction if brakes were used within a distance of 56m.
Solution: Using the work formula: W = F_{k} x ;  112,000 J = F_{k} (56m) ; F_{k} =  2000N.
Chapter 6 Test Yourself 1:
1) work is the product of (a) force and displacement (b) force and parallel displacement (c) parallel force and displacement (d) b and c. click here
2) The usual units for work is (a) Newton (b) Nm and lbft (c) kgf.
3) Joule is another name for (a) Nm (b) lbft (c) kgm/s^{2}. click here
4) In Example 2, if the θ = 40.0°, then the parallel force is (a) F (cos40.0) (b) F (sin 40.0) (c) F (tan40.0).
5) The work done by a force F is maximum if the direction of F is (a) normal to displacement (b) opposite to displacement (c) parallel to displacement. click here
6) The work done by force F through a displacement d is (a) Fd (b) F (dcosθ) (c) (Fcosθ)d (d) b & c.
7) Energy is defined as (a) the ability to apply force (b) the ability of doing work (c) work done per unit of time.
8) Energy is expressed in the same units as (a) mass (b) work (c) force. click here
9) Gravitational potential energy is the energy an object has due to its (a) speed (b) velocity (c) elevation.
10) With respect to the ground, the higher we take an object the greater its (a) mass (b) kinetic energy (c) potential energy becomes. click here
11) The kinetic energy (K.E.) of an object depends on its (a) elevation (b) its velocity (c) speed (d) b & c.
12) When a car is driven to the top of a hill, there is an increase in its (a) K.E (b) P.E (c) weight.
13) Potential energy (P.E.) is equal to (a) Mh (b) Mv^{2} (c) Mgh. click here
14) Gravitational potential energy makes more sense only when we are (a) in the vicinity of a planet (b) in the outer space (c) neither a, nor b.
15) Kinetic energy (K.E.) is equal to (a) (1/2)MV (b) MV^{2} (c) (1/2)MV^{2}. click here
16) When an object is lowered, its potential energy (a) increases (b) remains the same (c) decreases.
17) When an object moves at constant velocity, it K.E. (a) increases (b) decreases (c) remains unchanged.
18) A car in neutral with a certain initial velocity, can go up a hill by (a) gaining K.E. and losing P.E. (b) gaining P.E. and losing K.E. (c) none of a or b. click here
19) When the speed of a car is doubled, it K.E. (a) doubles (b) triples (c) quadruples.
20) When the speed of a car is tripled, it K.E. (a) triples (b) becomes 9 times greater (c) quadruples.
WorkKinetic Energy Theorem:
This theorem simply states that " the work done by the net force acting on a mass is equal to the change in the kinetic energy of that mass."
The mathematical formula for this theorem is: (ΣF) x = Δ(K.E.).
(ΣF) x = (K.E.)_{f }  (K.E.)_{i }.
The previous two examples combined show an application of this theorem. In Example 8, the change in the kinetic energy of the car was calculated. In Example 9, the work done by the net force was involved. The change in K.E. were set equal to the work done by force of friction, from which the force of friction was calculated. This shows that when energy is lost in one form, it appears in another form. One side of this formula calculates the work done by the net force ( Σ F ) within a distance x. The other side shows the change in the K.E.. Both sides are expressed in units of energy, however.
Example 11: A 900.kg car traveling at 15m/s changes its speed to 25m/s in a distance of 50.m due to a net force. Calculate (a) the net force, and (b) the engine force if the frictional forces add up to 1400N.
Solution: (a) Using the WorkK.E. theorem, we get:
(ΣF) ∙ x = (K.E.)_{f }  (K.E.)_{i} ; ΣF (50.m) = (1/2)(900)(25)^{2} _{ } (1/2)(900)(15)^{2} .
ΣF = 3600N.
(b) ΣF = F_{e}  F_{f} ; 3600N = F_{e}  1400N ; F_{e} = 5000N.
Example 12: A boy pushes a 45.0kg sled (including his friend in it) from rest for a distance of 12.0m with a horizontal and constant force of 165.0N. The frictional force between the sled and snow is 55.0N. Calculate (a) the work done by the boy on the sled, (b) the work done by the frictional force on the sled, (c) the work done by the net force on the sled, and (d) the speed of the sled at the end of the 12.0m distance.
Solution: (a) W_{boy} = F_{ll} x ; W = (165 N)(12.0 m) = 1980 J.
(b) W_{friction} = F_{k}x ; W = (55.0 N)(12.0 m) = 660. J. (Work of friction is negative).
(c) W_{net} = F_{net} x ; (ΣF) x = (165N  55.0N)(12.0 m) = 1320 J
(d) (ΣF) x = (K.E.)_{f }  (K.E.)_{i}
1320 J = (1/2)(45.0kg)V_{f}^{2}  (1/2)(45.0kg)(0)^{2 } ; V = (+/) 7.66 m/s.
The Law of Conservation of Energy:
This law states that "Energy is conserved. It is neither created nor destroyed. It converts from one form to another". Of course, this is true from the Universe point of view. From our point of view, when energy is converted to heat via friction, it is a loss. For example, when we push a heavy crate up a ramp (or incline) onto a truck, we do some work. Part of the work stores as P.E. in the elevated crate, and the remaining work converts to heat due to friction. It is not difficult to experimentally verify that
the work done on the crate = the P.E. stored in the crate + the work consumed for friction.
From our point of view, the work consumed for friction is a loss. From the Universe point of view, the work consumed for friction returns to nature in the form of heat and there is no loss.
Conservation of Mechanical Energy:
In the absence of frictional energy losses, the total mechanical energy of a system remains constant. We can always write down an energy balance for a system that goes under a certain process in which mechanical energy changes from one form to another, even if some energy is lost due to friction. The following examples apply the law of conservation of energy to a system or an object. Conservation of energy requires total energy to remain constant.
Example 13: In the figure shown, neglecting friction, find the speed of the 750kg car at the bottom of the hill. Suppose the car is put in neutral and starts from rest from the top of the hill.
Conservation of mechanical energy requires that Total Energy at A = Total Energy at B. (K.E.)_{A} + (P.E.)_{A} = (K.E.)_{B} + (P.E.)_{B}. (1/2)MV_{A}^{2} + M g h_{A} = (1/2)MV_{B}^{2} + M g h_{B}. Since V_{A} = 0 and also h_{B} = 0, we get: M g h_{A} = (1/2)MV_{B}^{2}. M's cancel and we get: g h_{A} = (1/2)V_{B}^{2}. 2 g h_{A} = V_{B}^{2} . V_{B} = 30. m/s.

Example 14: In the figure shown, if 120,000J of energy is consumed by frictional forces, find the speed of the 750kg car at the bottom of the hill. Suppose the car is put in neutral and starts from rest from the top of the hill.
In this case, part of the available energy will
be wasted by friction. Conservation of mechanical energy requires that Total Energy at A  W_{friction} = Total Energy at B. (K.E.)_{A} + (P.E.)_{A}  W_{friction }= (K.E.)_{B} + (P.E.)_{B}. (1/2)MV_{A}^{2} + M g h_{A} 120,000J = (1/2)MV_{B}^{2} + M g h_{B}. Since V_{A} = 0 and also h_{B} = 0, we get: M g h_{A} 120,000J = (1/2)MV_{B}^{2} . Not every term has M; therefore, M's do not cancel. (750)(9.8)(45) J  120,000 J = 0.5(750kg)V_{B}^{2} . 562(m/s)^{2} = V_{B}^{2}. V_{B} = 24 m/s.

Example 15: In the figure shown, find (a) the work done by the frictional force on the car as it coasts down the hill in neutral, (b) the energy loss due to friction, and (c) the speed of the 750kg car (V_{B}) at the bottom of the hill.
(a) The work done by friction is W_{friction} = F_{k}x or, W_{friction } = (300N)(400m) = 120,000J. (b) 120,000J (c) Conservation of mechanical energy requires that Total Energy at A  W_{friction} = Total Energy at B. (K.E.)_{A} + (P.E.)_{A}  W_{friction}= (K.E.)_{B} + (P.E.)_{B}. (1/2)MV_{A}^{2} + M g h_{A} 120,000J = (1/2)MV_{B}^{2} + M g h_{B}. Since V_{A} = 0 and also h_{B} = 0, we get: M g h_{A} 120,000J = (1/2)MV_{B}^{2}. (750)(9.8)(45) J  120,000 J = 0.5(750)V_{B}^{2}. 562(m/s)^{2} = V_{B}^{2} ; V_{B} = 24 m/s. 
Assume 2 significant figures on all numbers. 
A Good Link to Try: http://www.mhhe.com/physsci/physical/giambattista/roller/roller_coaster.html .
Power:
Example 16: An electric motor is capable of delivering 7.2 Million Joules of work in one hour and and 400. seconds. Find the power of the motor in watts, kilowatts, and hp.
Solution: P = W / t ; P = (7.2 x10^{6} J ) / (4000.s) = 1800 J/s = 1800 watts.
P = 1.8 kilowatts (kw) ; P = (1800 / 746 ) hp ; P = 2.4 hp.
Example 17: Calculate the amount of work or energy that a 4.50hp electric motor can deliver in 10.0 hours.
Solution: P = 4.5hp = (4.50)(746 watts) = 3360 watts or ( J /s ).
P = W / t ; W = Pt ; W = (3360 J/s)(36,000s) = 1.21 x 10^{8} J.
Efficiency: When a device receives power from a source, it does not deliver all of it in the intended form and converts a portion of it to other undesired forms. This makes a device to be less than 100 percent efficient. For example, a typical car receives chemical energy in the form of gasoline. The purpose is to convert it to kinetic energy. The efficiency of this conversion is way below 100 percent. A good portion of the original combustion of gasoline in the cylinders coverts to undesirable heat that has to be transferred to the ambient. Only in the winter time part of it is used to heat up the interior of the car. Electric motors have much better efficiencies. They normally return more than 90% of the electric power delivered to them in useful power. However, the generation of electricity itself is not very efficient. Most power plants have efficiencies of under 50 percent. A good nuclear power plant is about 45% efficient. Early coal plants were hardly 20% efficient. Efficiency of a device is defined as the ratio of power output to the power input for that device.
Eff. = P_{out} / P_{in } .
Example 18: A 1.25hp electric pump with an efficiency of 92.5% is used to perform 1.40 Mega Joules of work. (a) How long does it take for the electric pump to do the job? (b) If it was 100% efficient, how long would it take?
Solution: A 1.25hp electric pump means it draws 1.25hp of electric power from the source. It generally cannot give back all of it. Make sure that you write the formulas with horizontal fraction bars.
Eff. = P_{out} / P_{in} ; 0.925 = P_{out} / (1.25hp) ; P_{out} = 0.925(1.25hp) = 1.16hp.
(a) P = W / t ; t = W / P ; t = (1.40 x 10^{6} J) / (1.16 x 746 watts ) = 1620 sec.
(b) P = W / t ; t = W / P ; t = (1.40 x 10^{6} J) / (1.25 x 746 watts ) = 1500 sec.
Example 19: Calculate (a) the necessary power to empty a swimming pool with dimensions (12m)(25m)(2.4m) by pumping its water to an average height of 3.5m in 2.5hrs. The mass density of water is 1000.kg/m^{3}. ( Hint: The power you calculate is the power that the electric pump has to deliver to water). If the efficiency of the electric pump used is 88%, calculate (b) the electric power that the electric motor pulls from the power source.
Solution: V = w l h ; V = 720m^{3}.
Since density r = M / V ; therefore, M = r V; M = (1000 kg/m^{3})(720 kg) = 720,000kg.
(a) P = W / t ; W = F x, where F = the weight of water, and x = height.
P = (F x) / t ; P = ( Mg x) / t ; P = (720,000kg )( 9.8 m/s^{2} )( 3.5m) / (2.5 X 3600s) = 2700 watts.
P = 2.7kw ; P = (2700 / 746) hp = 3.6 hp.
(b) Eff. = P_{out} / P_{in} ; 0.88 = 2700watts / P_{in} ; P_{in} = 2700watts / 0.88 = 3100 watts.
Another Version of Power Formula:
The formula P = W / t can be written as P = (F x) / t because W = Fx. As we know ( x / t ) is the same as ( v ); therefore,
P = F v
This means that power is equal to force times velocity (if both of force and velocity are in the same direction). This concept is very familiar. Suppose you want to drive your car at a constant 3000rpm. If you are driving up a hill at 3000 rpm, you must gear down and drive at a lower velocity because you need a greater force between the tires and the ground. If you are driving on a flat and horizontal road, 3000rpm can give your vehicle a higher velocity because a smaller force is needed between the tires and the road on a horizontal road. As you see the product of velocity and force is a constant if a constant power is available. When P is constant, the product F∙v is constant. When a greater F is needed, a smaller v is attainable and vice versa.
Example 20: A car is using 56hp of its total power. When it is going up a steep hill, a force of 4200N is needed for constant velocity motion of it up that hill. When it is traveling along a level and horizontal road, a force of 1200N is needed for its constant velocity motion. Find the velocity of the car in each case.
Solution: P = (56)(746watts) = 42000watts.
(a) P = F v ; v = P / F ; v = 42000watts / 4200N = 10. m/s. ( Convert to km/h and mi/h).
(b) P = F v ; v = P / F ; v = 42000watts / 1200N = 35. m/s. ( Convert to km/h and mi/h).
Chapter 6 Test Yourself 2:
1) According to Work K.E. theorem, the work done on mass M by the net force within a distance x is equal to (a) the change in the K.E. of mass M (b) the change in the potential energy of mass M (c) change in the elevation of mass M.
2) When a car moving at velocity V_{1} uses an engine force F_{e} within a distance x to change its velocity to V_{2}, the work done by the engine force is (a) F_{e }x (b) V_{2}^{2}  V_{1}^{2} (c) 1/2MV_{2}^{2}. click here
3) In Question2, if the friction force acting on the car is F_{f}, the work done by the friction force on the car is (a) V_{2}^{2}  V_{1}^{2} (b) 1/2MV_{2}^{2} (c)  F_{f }x. click here
4) The work done by the net force acting on the car in Question 2 is (a) [F_{e}  F_{f}] x (b) MV^{2} (c) both a & b.
5) The change in the K.E. of the car in Question 2 is (a) (1/2)MV_{2}^{2}  (1/2)MV_{1}^{2} (b) (1/2)M [V_{2}^{2}  V_{1}^{2}] (c) both a&b.
Problem: A 1250kg truck is moving at 15.0m/s along the positive xaxis. The driver speeds it up to 25.0m/s within a distance of 125m. The frictional forces add up to 2800N. Use the WorkK.E. theorem to answer the following questions:
6) The change in the K.E. of the truck is (a) 6250J (b) 250,000J (c) 500,000J. click here
7) The work done on the truck by the net force acting on it is (a) 0 (b) Ma (c) 250,000J.
8) The net force acting on the truck in the xdirection is (a) 5000N (b) 2000N (c) Mg.
9) The engine force is (a) 4800N (b) 800N (c) 5880N. click here
Problem: A 750kg car traveling at an initial speed of V_{i} in the positive xdirection is brought to stop in a distance of 50.0m by applying brakes. The magnitude of the brakes force is 7500N. Note that when brakes are applied, usually the engine force drops to zero ( gas pedal is released). Use the WorkK.E. theorem to answer the following questions:
10) The net horizontal force acting on the car during the stopping process is (a) 7500N (b) 7500N (c) Mg.
11) The work done by the net force during the stopping process is (a) 375000J (b) 750000J (c) 375000J.
12) The change in the K.E. of the car is (a) 1/2(750)[0^{2}  V_{i}^{2 } ] (b) 1/2(750)[V_{i}^{2 } 0^{2}] (c) 375V_{i}^{2} (d) a & c.
13) Applying the WorkK.E. theorem results in the equality of 375000J = 375V_{i}^{2}. (a) True (b) False
14) The calculated value for V_{i} is (a) 100 m/s (b) 10 m/s (c) 31.6 m/s. click here
15) According to the law of conservation of energy, when we leave lights on without any use no energy is wasted (a) from our point of view (b) from the Universe point of view (c) neither a, nor b.
16) When you use brakes to stop your car, the energy that appears as heat in the brakes is (a) completely wasted and it is hard for us to get it back (b) not really wasted when viewing it from a Universal perspective (c) both a and b.
17) Conservation of Mechanical Energy Theorem involves (a) K.E. only (b) K.E. & Gravitational P.E. only (c) the energies in (b) plus Elastic (Spring) P.E. as well. click here
Problem: Suppose you are driving along a straight and horizontal highway in a 650kg car (including your own mass) and your speed is 30.0m/s. The road becomes uphill and you suddenly become curious as to how high up the hill your car might go if you put it in neutral and let it come to stop on its own. Suppose that you have a good estimate of energy losses due to frictional forces and that is 85500J. Answer the following questions:
18) Before starting the uphill and with respect to the horizontal road, your car has (a) P.E. only (b) K.E. only (c) both P.E. and K.E. click here
19) As your car goes up the hill it (a) loses P.E. and gains K.E. (b) loses K.E. and gains elastic energy (c) loses K.E. and gains P.E..
20) In the absence of friction, at any intermediate point before your car comes to stop, the total energy (P.E. + K.E.) of your car would be (a) constant (b) increasing (c) decreasing. click here
21) In the absence of energy loss due to friction, the P.E. of your car at the highest point it can go, will be equal to (a) its K.E. halfway up the hill (b) its K.E. at the starting point (c) neither a, nor b.
22) The car's K.E. at the start of the hill is 292500J. (a) True (b) False
23) The P.E. at the top of the hill and in the absence of friction would have been 292500J. (a) True (b) False
24) The actual P.E. of your car at the highest point it can reach is (a) 292500J (b) zero (c) 207000J.
25) The formula that calculates P.E. is (a) P.E. = 1/2 Mgh (b) P.E. = Mg (c) P.E. = Mgh. click here
26) From 24 and 25, we may calculate the highest elevation your car reaches as (a) 45.9m (b) 32.5m (c) 318.5m.
27) In the absence of friction, the highest elevation it could reach would be (a) 45.9m (b) 318.5 (c) 400m.
Power
28) Power is defined as (a) the rate of doing work (b) the work done per unit of time (c) the energy generation per unit of time (d) a, b, and c. click here
29) The power of a motor that lifts 4500N of weight to a height of 25m in 15s is (a)7500 watts (b)7.5 kilowatt (c) both a & b.
30) One hp is equal to (a) 746 watts (b) 550 (lbft) /s (c) both a & b.
31) The normal power that a car can generate is 120kw (120,000 watts). At normal power, in going up a hill, a force of 15000N must be delivered to the tires to maintain a constant velocity. The constant velocity at which the car goes up the hill is (a) 8.0m/s (b) 12m/s (c) 16m/s. click here
32) What is the constant speed of going up a milder slope if the necessary force applied to the tires is 12000N? (a) 20m/s (b)18m/c (c) 10.m/s.
33) What is the constant speed of going up a more milder slope if the necessary force applied to the tires is 6000N? (a)25m/s (b)20m/c (c) 30.m/s.
34) On a horizontal road where the necessary force applied to or by the tires is 3000N, the cars speed at normal power usage is (a) 50m/s (b) 60m/s (c) 40 m/s. click here
35) The necessary power that an electric pump must deliver to water is 2700 watts. The pump is 90% efficient. The electric power that it pulls from the electric socket is (a) 2430 watts (b) 3000watts (c) neither a, nor b.
Problems:
1) A constant net force has done 3460J of work in pushing a 30.0kg block on a uniform horizontal floor through a distance of 26.4m. Find (a) the net force, (b) the friction force if μ_{k} = 0.425, and (c) the applied force.
2) A truck that weighs 23520N is moving east on a straight and horizontal road at a constant speed of 13.5m/s. Find (a) its mass, and (b) the net force acting on it. Its driver decides to accelerate the truck and therefore changes the net force to 4480N. If this net force is kept constant for a distance of 125m, find (c) the final speed of the truck. If the coefficient of road friction is μ_{k} = 0.0961, calculate (d) the friction force of the road. If air resistance is 1260N, find the engine force during the acceleration period. g = 9.80m/s^{2}.
3) What horizontal force can push a 40.0kg block placed on a horizontal floor (a) at constant velocity to the right, and (b) at an acceleration of 1.65m/s^{2} to the right? (c) For each case, calculate the work done by the pushing force within a distance of 12.0m. Let μ_{k} = 0.275 and g = 9.81m/s^{2}.
4) A 1680kg spacecraft is traveling at a speed of 72,000 km/h along a straight path in outer space. Calculate (a) its kinetic energy, (b) its potential energy, and (c) the net force acting on it. Its rockets are then turned on for a distance of 2450km and as a result it receives a constant thrust of 25.0 kN within this distance. Calculate (d) the net work done by the rockets, and (e) its final speed.
5) Neglecting friction, determine how far up the hill will the 862kg car go due to its initial speed at A, if its engine is turned off at A and put in neutral.


6) If 76,300J of energy is used up by friction from A to B, determine how far up the hill will the 862kg car go due to its initial speed at A, if its engine is turned off at A and put in neutral.


7) In the figure shown, the effective force of friction along the uphill is 3000N. Determine (a) how far, x, up the hill will the 862kg car go due to its initial speed at A, if its engine is turned off at A and put in neutral. (b) Calculate the energy loss due to friction.


8) How long does it take for a 90.0% efficient 3.00hp electric pump to empty a (14.0m )( 25.0m )( 2.00m) swimming pool, if its water is to be pumped to an average height of 5.00m? 1hp = 746 watts. g = 9.81m/s^{2}.
9) The power delivered by a car at 3000 rpm is 36.2 hp. The needed force in the tires of this car on a horizontal road is 1350N, on a road with a mild slope, 1800N, and on a road with a steeper slope, 2700N. Calculate, the speed at which this car travels along each of the roads, respectively, if its power is kept constant.
10) A 90.0% efficient electric motor delivers 2.70 hp of effective power to a device. Find (a) the power it draws from the electric outlet (a) in hp, (b) in watts, and (c) in kilowatts. If this motor is used for 15.0 hours, (d) how much electric energy (in kwh) does it consume? (e) If the price of electric energy is 7.25 ¢ per kwh, calculate the cost.
Answers: 1) 131N, 125N, 256N 2) 2400kg, 0, 25.5m/s, 2260N, 8000N. 3) 108N, 174N, 1300J, 2090J
4) 3.36x10^{11} J, 0, 0, 6.125x10^{10}J, 78,300km/h 5) 35.8m 6) 26.8m 7) 41.9m, 126000J
8) 4h, 44min. 9) 20.0m/s, 15.0m/s, 10.0m/s. 10) 3.00hp, 2240watts, 2.24kw, 33.6kwh, $2.44