A 12kg block is held in place against the spring by a 35N horizontal external fo

Question

A 12kg block is held in place against the spring by a 35N horizontal external force. The external force is removed, and the block is projected with a velocity v1=1.2 m/s when it separates from the spring, as shown in the figure. The block descends a ramp and has a velocity v2 = 2.1 m/s at the bottom of the ramp. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction between the block and the rough surface is 0.26. The velocity of the block is v3 = 1.4 m/s at C. The block moves on to D, where it stops.

By how many centimeters was the spring initially compressed? = 0.49 m

What is the height of the ramp? = 0.15 m

How much work is done by friction between points B and C? = 14.7 J

What distance does the block travel between points B and D? = 0.87m

Step by step how to solve these problems?

Explanation / Answer

energy stored in spring = k x^2 /2 = m v1^2 /2

= 12 x 1.2^2 /2 = 8.64 J

so k x^2 /2 = 8.64

k x^2 = 17.28 ....(i)

F = kx

k x = 35 .... (ii)

i / ii => x = 17.28 / 35 = 0.49 m ........Ans

Applying work energy theorem,

Work done by gravity = change in KE

m g h = m (v2^2 - v1^2 ) / 2

2 x 9.8 x h = 2.1^2 - 1.2^2

h = 0.151 m ..........Ans

Work done by friction = change in KE

W = 12 (1.4^2 - 2.1^2) /2 = - 14.7 J ........Ans

from B to D.

W = 12(0^2 - 2.1^2) / 2 = - 26.46 J

f = uk N = uk m g

work done by friction, W = - uk m g d

- 26.46 = - 0.26 x 12 x 9.8 x d

d = 0.87 m .......Ans

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