menu search
Feel free to answer or ask any questions.
If you are new here please see how to use or FAQ.

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 initally at rest in its equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is :

(1) $\pi F\over\sqrt{mk}$

(2) $2F\over\sqrt{mk}$

(3) $F\over\sqrt{mk}$

(4) $F\over\pi\sqrt{mk}$

thumb_up_off_alt 3 like thumb_down_off_alt 0 dislike

1 Answer

Best answer

Ans: (3) $F\over\sqrt{mk}$

Sol: Maximum speed is at mean position (equilibrium). F = kx

$x={F\over k}$ 

$W_F + W_{sp} = \Delta KE$

$F(x) – {1\over2} kx^2={1\over2}mv^2-0$

$F\left({F\over k}\right) – {1\over2} k\left({F\over k}\right)^2={1\over2}mv^2$

$\implies v_{max} = {F\over\sqrt{mk}}$

thumb_up_off_alt 2 like thumb_down_off_alt 0 dislike
Welcome to Jee Neet QnA, where you can ask questions and receive answers from other members of the community.

1.1k questions

772 answers


79 users