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

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.

Work, Energy and Power & Circular Motion

Integer / Numeric Answer Type