Conservation of Energy
The objective is to verify the law of conservation of mechanical energy with frictional losses present.
An incline, a wooden box, a pulley with attachments, a 1-meter piece of string, a mass scale, graphing paper, a set of weights, a weight hanger, and a scientific calculator
For an inclined plane as shown in Fig. 1, the P.E. at position 2 exceeds the P.E. at position 1 by the amount M1gh where M1 is the total mass (box + added mass) placed on the incline and h is the change in the elevation of the box.
M1gh is the work done on the box in going from position 1 to position 2 if no friction is involved; however, there is friction between the box and the incline and more work must be done to elevate the box from position 1 to position 2. Mass M2 provides the necessary work or energy. If M2 is great enough to pull the box up the hill while M2 itself moves down under the influence of gravity, and the process occurs at a constant velocity, we may then accept that
Energy loss by M2 = Energy gain by M1 + Energy consumed by friction.
This can mathematically be written as:
M2 g S = M1 g h + Wf (1)
in which h and Wf as well as μ, the coefficient of friction, must be determined.
Calculation of h: To determine h, refer to the right triangle shown in Fig. 2:
Measurement of μ:
To measure μ, tilt the incline until the box slides down at a constant velocity. Measure that particular angle, θk, at which constant-velocity sliding occurs and use it to find μk from the formula (For proof refer to Experiment 5):
μk = tan θk
Calculation of Wf :
To calculate the work or energy consumed by friction, we need to determine the force of kinetic friction, Fk , and then multiply it by S, the displacement. Let's denote the Work done by friction by the upper case w as Wf. Lower case w is used to denote the weight of the box or M1g.
Fk is given by Fk = μk N, where N can be found from the figure on the right.
N = w cos θ. We may write:
Wf = Fk S
= μk N S = [μk w cosθ] S or,
Wf = μk M1 g (cos θ) S.
w, the weight, can be replaced with F┴ and F║ . F┴ is a measure of how strong the box presses against the incline. Note that F┴ must be equal to N for equilibrium in the normal to the surface direction. Since F┴ = wcosθ ; therefore, N = wcosθ as well.
Finally, Eq. 1 becomes:
M2 g S = M1 g S (sin θ) + μk M1 g (cos θ) S. (2)
Eq. 2 is the energy balance equation. If SI units are used, both sides will be in Joules. Although S can be cancelled out from both sides, but we deliberately keep it in there and let its value be 0.300m, for example. This will result in units of Joules if M1 and M2 are converted from grams to kg. We will then be comparing the number of joules we get on the left side with the number of joules on the right side of this equation.
Note that since the experiment is to be performed at a constant velocity, the initial and final velocities will be equal and therefore there will not be any change in the K.E. of the system. That is why no mention is made of K.E. in Eq. 2.
With the box (empty or with some weights in it) on the incline, tilt the incline to angle θk where motion at constant velocity can be observed. Repeat this procedure 3 times for the empty box and 3 times with 300 grams of mass in it and each time determine a value for μk by using the appropriate equation. Find an average value for μk based on the 6 angles you measure.
μk = tan θk
The Main Experiment:
1) Set up the incline at angle θ between 20o to 30o. With 300gr in the box, find M2 (the hanging mass) such that the box slides up the incline at a slow constant velocity. Record the values of θ, M1 (box +300gr), and M2 (weight hanger + added weights on it) in the first row of Table 1 in correct units.
2) Assuming a displacement of say (S = 0.300m or so) in your calculations, use Eq. 2 and calculate the values of each side of the equation, separately, and see how close in value they turn out to be. Record the calculated values (in Joules) in Table 1.
3) Calculate a % difference between the right and left values by using the following equation and record it in Table 1.
4) Repeat the above steps for M1 = 400 grams and the same θ.
5) Change θ to another angle between 20o and 30o, and repeat the experiment once for M1 = 300 grams and once for M1 = 400 grams again.
6) Record all results in Table 1. In each case show the calculated percent difference.
Given: g = 9.81 m/s2, 20.0o < θ < 30.0o, M1 = 300gr. & 400gr.
Left side of Equation
|# of Joules
Right side of Equation
Comparison of the results:
Show the percent difference formula used as well as the calculated values of percent difference for all cases.
State your conclusions of the experiment.
Provide a discussion if necessary.