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A 10 kg box is traveling 5 m/s down a 4 meter inclined plane

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A 10 kg box is traveling 5 m/s down a 4 meter inclined plane that is at an angle of 37 degrees. The coefficient of friction is .4. What is the change in potential energy, the work done by the friction force and the speed of the box before it reaches the end of the inclined plane.

A 10 kg box is traveling 5 m/s down a 4 meter inclined plane that is at an angle of 37 degrees. The coefficient of friction is .4. What is the change in potential energy, the work done by the friction force and the speed of the box before it reaches the end of the inclined plane.

Change in potential energy = mgh = mg*4sin 37

= 10*9.8*4*0.6018 = 235.91Joules.

Work done by frictional force = μmgcosθ*distance travelled

= 0.4*10*9.8*cos37*4

= 125.226J

If velocity change at end is v, we have 0.5mv^{2}= mgh

Or v = sqrt(2gh) = sqrt(2*9.8*4sin37) = 6.868m/sec

Hence, velocity at end = 5+6.868 = 11.868m/sec

Kindly ACCEPT the answer. BONUS and POSITIVE feedback will be appreciated.

Frictional force can appear in many situations. In one situation, it may be conservative and in the other situation,it may be non-conservative. For solving this problem, it is not required to know whether frictional force is conservative or not.