A 12.0 kg block is draggedover a rough, horizontal surface by a 95.0 N force act
ID: 1753347 • Letter: A
Question
A 12.0 kg block is draggedover a rough, horizontal surface by a 95.0 N force acting at 20.0° above thehorizontal. The block is displaced 6.00 m, and the coefficient of kineticfriction is 0.300. (a) Findthe work done on the block by the 95.0 N force.J
(b) Find the work done on the block by the normal force.
J
(c) Find the work done on the block by the gravitational force.
J
(d) What is increase in internal energy of the block-surface systemdue to friction?
J
(e) Find the total change in the block's kinetic energy.
J (a) Findthe work done on the block by the 95.0 N force.
J
(b) Find the work done on the block by the normal force.
J
(c) Find the work done on the block by the gravitational force.
J
(d) What is increase in internal energy of the block-surface systemdue to friction?
J
(e) Find the total change in the block's kinetic energy.
J
Explanation / Answer
a) work done by 95N force = Fd cos = 95(6)(cos200) J b) work done by the normal force = 0 (since there is nodisplacement in the normal direction). c) work done by the gravitational force = 0 (since there is nodisplacement in the vertical direction). d) increase in internal energy due to friction = work done byfriction = (frictional force)(displacement). friction = N N + F sin = mg friction = (mg - F sin ) increase in internal energy due to friction = (mg - F sin)(6) >>> you can compute this since you know ,m, F and . e) change in KE = net work done = work done by 95N force - workdone by friction = answer to part (a) - answer to part (d) hope this helps!
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