Question

The 52 kg block shown starts from rest at position A and slides down the inclined plane to position B. When the block reaches position B, a 383 N horizontal force is applied. The block comes to a complete stop at position C. The coefficient of friction between the block and the plane is $mu = 0.15$. P = 383 N

          The 52 kg block shown starts from rest at position A and slides down the inclined plane to position B. When the block reaches position B, a 383 N horizontal force is applied. The block comes to a complete stop at position C. The coefficient of friction between the block and the plane is $mu = 0.15$.
P = 383 N

The 52 kg block shown starts from rest at position A and slides down the inclined plane to position B. When the block reaches position B, a 383 N horizontal force is applied. The block comes to a complete stop at position C. The coefficient of friction between the block and the plane is mu = 0.15.
P = 383 N

Added by Melissa K.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Ivan Kochetkov

06/10/2025

Step 1

The force applied at position B is 383 N, and it brings the block to a stop at position C. This means that the work done by the force is equal to the change in kinetic energy of the block. The work done by the force is given by the equation W = F * d, where F Show more…

Show all steps

Thanks for your feedback!

1. Find the distance between positions B and C and the velocity at position B The 52 kg block shown starts from rest at position A and slides down the inclined plane to position B. When the block reaches position B, a 383 N horizontal force is applied. The block comes to a complete stop at position C. The coefficient of friction between the block and the plane is μ = 0.15.