Steps important! In the figure, a 4-kg block is released from rest and accelerat

Question

Steps important! In the figure, a 4-kg block is released from rest and accelerated by a compressed spring of spring constant 900 N/m. When the block leaves the spring at the spring's relaxed length, it has a kinetic energy of 8 J. It then travels over a horizontal floor where a kinetic frictional force of 6 N acts on the block. The frictional force stops the block in distance D. The system consists of the block and spring. What is the elastic potential energy of the system when the block is released from rest? There is no friction while the block is in contact with the spring. (8J) Determine the distance the spring is compressed when the block is released from rest. (2/15 m) How far does the block slide after leaving the spring? In other words, find the distance D. (4/3 m)

Explanation / Answer

(A) Spring potential energy converts into KE of blcok.

when spring is in relaxed position then all the spring PE is transferred as KE of block.

hence using energy conservation,

spring PE = KEmax = 8J ......Ans

(B) spring PE = k x^2 /2

8 = (900)(x^2) /2

x = sqrt[16 / 900] = 4/30 = 2/15 m

(C) using work - energy theorem,

Work done by friction = change in KE

- f .D = Kf - Ki = 0 - 8

f.D = 8

(6)(D) = 8

D = 4/3 m

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