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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.
Change in potential energy = mgh = mg*4sin 37
= 10*9.8*4*0.6018 = 235.91Joules.
Work done by frictional force = μmgcosθ*distance travelled
If velocity change at end is v, we have 0.5mv2= mgh
Or v = sqrt(2gh) = sqrt(2*9.8*4sin37) = 6.868m/sec
Hence, velocity at end = 5+6.868 = 11.868m/sec
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