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Question

Physics Question on work, energy and power

A block of mass mm, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant kk. The other end of the spring is fixed, as shown in the figure. The block is initally at rest in its equilibrium position. If now the block is pulled with a constant force FF, the maximum speed of the block is :

A

πFmk\frac{\pi F}{\sqrt{mk}}

B

2Fmk\frac{ 2 F}{\sqrt{mk}}

C

Fmk\frac{F}{\sqrt{mk}}

D

Fπmk\frac{F}{ \pi \sqrt{mk}}

Answer

Fmk\frac{F}{\sqrt{mk}}

Explanation

Solution

Maximum speed is at mean position
(equilibrium). F=kxF = kx
x=Fkx = \frac{F}{k}
WF+Wsp=ΔKEW_{F} + W_{sp} = \Delta KE
F(x)12kx2=12mv20F\left(x\right) - \frac{1}{2} kx^{2} = \frac{1}{2} mv^{2} -0
F(Fk)12k(Fk)2=12mv2F\left(\frac{F}{k}\right)- \frac{1}{2} k \left(\frac{F}{k}\right)^{2} = \frac{1}{2} mv^{2}
vmax=Fmk\Rightarrow v_{max } = \frac{F}{\sqrt{mk}}