Question 338: Work, Energy and Power & Circular Motion

Question

A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The
coefficient of friction between the block and the floor is 0.1. Initially the
block is at rest and the spring is un-stretched. An impulse is given to the
block as shown in the figure. The block slides a distance of 0.06 m and
comes to rest for the first time. The initial velocity of the block in m/s is
V = N/10. Then N is
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l2jfj8nb/26b37588-c6bf-4721-b3f4-438fcf9c4ad0/b10ba970-c72d-11ec-9e4a-bf17a41218c9/file-1l2jfj8nc.png?format=png)

Accepted Answer

Loss of kinetic energy $ = {1 \over 2}m{V^2}$
Work done against friction $ = \mu mgx$
Gain in potential energy $ = {1 \over 2}k{x^2}$
From work-energy principle,
${1 \over 2}m{V^2} = \mu mgx + {1 \over 2}k{x^2}$
$ \Rightarrow {1 \over 2} 	imes 0.18 	imes {V^2} = 0.1 	imes 0.18 	imes 10 	imes 0.06 + {1 \over 2} 	imes 2 	imes {(0.06)^2}$
$ \Rightarrow V = 0.4 = {4 \over {10}}$ ms
$-$1
. Hence, N = 4.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l4k3c275/e709e373-7712-4094-8945-9e512521b0cb/3341d910-ef23-11ec-af85-e5c347851f91/file-1l4k3c276.png?format=png)

Work, Energy and Power & Circular Motion

Integer / Numeric Answer Type