JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Work, Power & Energy

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

(A) $\frac{2 F}{mk}$
(B) $\frac{F}{π mk}$
(C) $\frac{π F}{mk}$
(D) $\frac{F}{mk}$ ✓ Correct

Answer: (D)

Solution

The block keeps speeding up only while the applied force beats the spring force, so maximum speed occurs exactly where they balance,

k x = F

, i.e.

x = \frac{F}{k}

. Apply the work-energy theorem from the start to this point: the applied force does work

F x

while the spring stores

\frac{1}{2} k x^{2}

, so

F x - \frac{1}{2} k x^{2} = \frac{1}{2} m v_{m a x}^{2}

. Putting

x = \frac{F}{k}

leaves

\frac{F ^{2}}{2 k} = \frac{1}{2} m v_{m a x}^{2}

, hence

v_{m a x} = \frac{F}{mk}

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Work, Power & Energy

Solution

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