Now, for the car of mass M moving down the incline at constant speed, we can sho

Question

Now, for the car of mass M moving down the incline at constant speed, we can show, form 2nd law of motion, that the force of friction f is given by f = M g sin theta - eta g where eta is the mass hanging from the other side of the pulley which makes the car move down up. And finally the work done by friction becomes: Wf = f d Where d is the distance the car travels along the board. Determine the mass M of the car and record it. Set up the board and the car as shown in figure 1 with an incline angle of 30 theta with the table. Position the car near the bottom of the incline and record its initial height h. Add enough weights of mass p on the weight hanger to make the car move up the incline at a very slow speed after given a slight tap. Record the initial height H of his weight. Measure also the distance d that the car travels along the board. When the car reaches near the top of the incline, record now its height h and that of the weight hanger H. Start removing weights from the weight hanger until the car starts moving down the incline at a slow speed after given a small tap. Record this mass eta. Repeat the above procedures for an incline angle of 60 degree. Compute the percent energy lost to friction for the car moving up the incline for both angles. For theta = 60 degree . Compute the work done by each force acting on the car. Should they add up to Zero Explain. For theta = 30 degree. Compute the normal force acting on each wheel of the car? Is friction in this experiment only due to the car? Explain. Show that, for the car moving up the incline, that the coefficient mu of friction is given by: mu = rho/M cos theta - tan theta

Explanation / Answer

forces on M T- Mgcos(theta) = Ma forces on n ng -T = na adding both ng-Mgcos(theta) = (M+n)a a = [ng-Mgcos(theta)]/(M+n) F = [ng-Mgcos(theta)]/n(M+n)

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