Question

А D z Problem 3. Work done by gravity and change in gravitational potential energy In problem the box was moving in a horizontal direction, and therefore no work was done by gravity. Here, we will analyze a situation where the force of gravity has some component that points along the direction of the displacement, and therefore there is non-zero work done by gravity on the system of interest Consider a box of mass 10 kg, initially at rest, which slides down a frictionless incline, as shown in the figure at right. The incline has a length of Sm and makes an angle of 20 with the horizontal. For parts A Ctake gravity to be outside the system, so that the force of gravity does work on the system A. Draw a free-body diagram of the box. Next to your diagram, show the direction of the displacement of the box B. What is the work done by each of the forces in your free-body diagram? Take care to include the correct sign for cach non-zero work (positive or negative) (1) W- W = d C. What is the kinetic energy of the box when it reaches the bottom of the incline? Use the work-energy theorem W = AK for this part. Now, let's do the problem again, but taking gravity to be part of the system. If the box starts at rest at the top of the incline then it will have initially gravitational potential energy stored in it, so to find the kinetic energy of the box at the bottom of the incline we need to use the more general form of the work-energy theorem: W = A + AU. D. Now that we are taking gravity inside the system, the only external force is the normal force. What is the total work done on the system as the block slides down the slope? 3

Answer 1

А) sm mal lokg da sm N m a displacement N=mg coso mgcoso mgsine B WG- (mg sino) a (10kgx: mis? Sin 28x5m = 171J WN- mg coso )d cos quo O: ДК W = WGT Ww=DK = KEF-KE; 111:5+0J= kiep-so's [REF KEF=171J SME KE po (at top ) k Ep: Kinetic SV energy at botton D W: ΔK+ ΔΟ DU= Up - U} = O Jit-171 J AK - KEG-KE; ורו OJ - RJ 1415 V MOP 11-5+ (-171.J.) - OJ WOJ