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To verify the law of conservation of mechanical energy with frictional losses present

An incline, a wooden box, a pulley with attachments,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. 2, the P.E. at position “2” exceeds the P.E. at position “1” by the amount M₁gh where M₁ is the total mass( box + 300. grams) 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 “1” to “2” if no friction is involved ; however, there is friction between the box and the inclined 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 constant velocity, we may then accept that

Energy delivered by M₂ = Energy consumed by M₁ + Energy consumed by friction

Where h and W_f as well as μ, or the coefficient of friction, must be determined.

To measure μ, tilt the incline until the box slides down at constant velocity. Measure the particular angle, θ_k, at which constant-velocity sliding occurs and use it to find μ_k from the formula (For proof refer to Experiment 5):

To calculate the work or energy consumed by friction, we need to determine the force of kinetic friction, F_k, and then multiply it by S, the displacement. Let's denote the Work done by friction by the upper case w as W_f. Lower case w is used to denote the weight of the box or M₁g.

F_k is given by F_k = μ_k N, where N can be found from the figure on the right.

N = w cos θ. We may write:

W_f = F_k S

= μ_k N S = [μ_k wcosθ] S ,or

W_f = μ_kM₁g (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.

Eqn. 2 is the energy balance equation. If SI units are used, both sides will have units of 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 constant velocity, the initial and final velocities will be equal and therefore there won’t be any change in the K.E. of the system. That is why no mention is made of K.E. in Equation (2).

With the box (empty or with some weights in it) on the incline, tilt the incline to θ_k 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 μ_k by using the appropriate equation. Find an average value for μ_k based on the 6 angles you measure.

Set up the incline for an angle θ between 20 to 30 degrees. With 300. grams in the box, find M₂ such that the box slides up the incline at constant velocity. Assuming a displacement of say (S = 0.300m or so), use Equation (2) and calculate the values on both sides of the equation and see how close in value they turn out to be.

Calculate a %difference between the right and left values by using the following equation:

Change θ and repeat the experiment once for M₁ = 300.grams and once for M₁ = 400. grams.

Calculation(s):

Provide the necessary calculations.

Provide the percent error formula used as well as the calculation of percent errors.