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Question

# A block of mass 2.0 kg at rest on an inclined plane of inclination 37∘ is pulled up the plane by applying a constant force of 20 N parallel to the incline. The force acts for one seconds. (a) Show that the work done by the applied force does not exceed 40 J. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is 40 J. (c) Find the kinetic energy of the block at the instant the force ceases to act. Take g=10 m/s2.

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Solution

## Force F = 20N .Mass m = 2.0 kgInitial velocity u = 0acceleration, a = 10m/s2t = 1 sec.Now Refer to the attachment, See the free body diagram of the block.Force works on the block:-Weight, W = mgW = 2 × 10W = 20N (which is Downward) Normal force N = mg cos37N = 20 × 0.80 N = 16 N. (perpendicular & upward to the plane )Here Applied Force, P = 20N (which is down along the plane)Now For Final Speed, We know the formula:-v = u + atv = 0 + 10 × 1v = 10 m/sthe Distance travelled s = ut + 0.5 at×ts = 0 + 0.5 ×10×1×1s = 5 m.Now,(a) So work done by the force of gravity in 1 sec. = F × d = 20 N × 5m = 100 J.(b) Here the weight act as downward, so distance travelled in downward. = 5 × sin37 = 5 × 0.6 = 3 m.so work done by gravity, = 20 N × 3 m = 60 J.(c) Now, work done by all the forces = change in Kinetic energy = 12m(v2−u2) = 0.5×2.0×(102−02) = 100 J.W.D by frictional force = work was done by all forces -( work was done by Normal force + work done by applied force + work done by gravity ) = 100 J - (100 + 60 +0 ) = 100 - 160 = -60 J

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