A 5.0 kg block at the top of a ramp is given a quick push, and thus it has an in

Question

A 5.0 kg block at the top of a ramp is given a quick push, and thus it has an initial speed of 1.5 m/s as it starts to slide down the ramp. The vertical height of the ramp is 8.0 meters, and its angle is 20 degrees with respect to the horizontal. The coefficient of kinetic friction between the block and the ramp is 0.20. Use g = 10 m/s2.

(a) What is the gravitational potential energy of the block at the top of the ramp? Assume the gravitational potential energy of the block is zero at the bottom of the ramp.

(b) What is the kinetic energy of the block at the bottom of the ramp?

(c) What is the speed of the block at the bottom of the ramp?

Explanation / Answer

a) PE= mgh = 5*10*80 = 400 J

b) By law of conservation of energy

Initial ME = Final ME

PEi + KEi= Efriction+ KEf

KEf = PEi - Efriction -KEi

KEf = mgh - uk*Fn*d -1/2mvi^2

KEf = mgh - uk*mgcosq*d -1/2mvi^2

d= h/sinq

KEf = mgh - uk*mgcosq*(h/sinq)-1/2mvi^2

c) from b)

KEf = mgh - uk*mgcosq*(h/sinq)-1/2mvi^2

1/2mvf^2 = mgh - uk*mgcosq*(h/sinq) - 1/2mvi^2

vf= sqrt[2 (gh - uk*gcosq*(h/sinq)-1/2vi^2)]

vf= sqrt[2 (10*8 – 0.20*10*cos20*(8/sin20)-1/2*1.5^2 )] = 8.36 m/s

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