<p>A box of 2-kg is released from rest from the top of an inclined surface of 37

Question

<p>A box of 2-kg is released from rest from the top of an inclined surface of 37 degrees travels down and moves along a horizontal surface. There is no friction on the inclined surface, but there is friction on the horizontal surface. (a) What is the speed of the box as it reaches the bottom? (b) Once the box reaches the bottom, it keeps moving on the horizontal surface where there is friction with kinetic coefficient 0.2. On the other side of the horizontal surface, there is a spring with k=100 N/m. The box compresses the spring by 1.5 m and it stops. Find the displacement of the box on the horizontal surface. (The incline surface is 3m long and the horizontal surface before the spring is 5m.)</p>
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Explanation / Answer

Calculate height of the inclined plane:

sin37 = h/3     h = 1.2m

(a) For this part we will use law of conservation of mechanical energy.Since there is no friction,therefore initial energy is equal to final energy.

Initial energy, Ei = mgh

Final energy, Ef = 1/2 mv2

Ei = Ef

mgh = 1/2 mv2

v = (2gh) = (2*9.8*1.2) = 4.85 m/s

(b) For this part the provided information is not appropriate. You provided lenghth of horizontal part 5m.

If this is the case then the spring can not be compressed by 1.5 m. The maximum compression of the spring in this case will be 0.28 m.

Solution: Let total distance travelled on horizontal surface = S

The total initail energy will be lost in doing work against friction.

Initial energy Ei = mgh = 2*9.8*1.2 = 23.52 J

Work done against friction W = mgS = 0.2*2*9.8*S = 3.92S

3.92S = 23.52

S = 6 m.

Total distance covered in horizontal surface is 6 m.

I hope this helps you.In case you ned further help please mail me I will be happy to reply.

Thank you.

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