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السؤال 2
A block slides on a rough horizontal surface from paint A to point B. A force (magnitude P -
2.0 N) acts on the block between A and B, as shown. If the kinetic energies of the block at A
and B are 5.0 J and 4.0 J, respectively and the work done on the block by the force of friction
as the block moves from A to B is -4.5 J what is the distance between point A and B?
40°
B​

1 Answer

3 votes

The work-energy theorem says that the total work done on the block is equal to the difference of its kinetic energies at points B and A. Then the total work done on the block is


W_(\rm total) = K_B - K_A = 4.0\,\mathrm J - 5.0\,\mathrm J = -1.0\,\mathrm J

Friction acts on the block to oppose its motion, so it does negative work on the block, -4.5 J.

The only other force acting on the block as it moves is the force P. Let
W_P be the work done by the force P. Then the total work done on the block is


W_(\rm total) = W_P + W_(\rm friction) \iff -1.0\,\mathrm J = W_P - 4.5 \,\mathrm J \implies W_P = \boxed{3.5\,\mathrm J}

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