a 20 kg box slides 45 m from the top down to the bottom ofan incline which is at

Question

a 20 kg box slides 45 m from the top down to the bottom ofan incline which is at an angle of 10 degrees to the horizontal. the box is stationary at the top of the slope and is accelerating down the slope at a rate of 1.0 m/s^2.

A. How much work is done by the force of gravity on the box?

B. How much work on the box is done by the frictional force?

C. What is the kinetic energy of the box at the bottom of the slope?

D. How much gravitational potential energy has the box lost while traveling down the slope?

PLEASE SHOW WORK

Thank you

Explanation / Answer

a ) 20*9.8*sin10 *45 = 1531.57 J

a = 1 = 9.8*sin10 - m*9.8*cos10

m= coefficient of friction

m= 0.073

b) work done by friction = 0.073*20*9.8*cos10*45 = 634.078 J

c) KE = 1531.57-634.078 = 897.492 J

d) loss in gravtational potential energy = 20*9.8*45*sin10 = 1531.57 J

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