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 :

Question

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 : (A) (π F/√mk) (B) ( 2 F/√mk) (C) (F/√mk) (D) (F/ π √mk)

Tardigrade Admin · Accepted Answer

Maximum speed is at mean position (equilibrium). F = kx x = (F/k) WF + Wsp = Δ KE F(x) - (1/2) kx2 = (1/2) mv2 -0 F((F/k))- (1/2) k ((F/k))2 = (1/2) mv2 ⇒ vmax = (F/√mk)

$\frac{π F}{mk}$

$\frac{2 F}{mk}$

$\frac{F}{mk}$

$\frac{F}{π mk}$

Maximum speed is at mean position (equilibrium). $F = k x$
$x = \frac{F}{k}$
$W_{F} + W_{s p} = Δ K E$
$F (x) - \frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} - 0$
$F (\frac{F}{k}) - \frac{1}{2} k (\frac{F}{k})^{2} = \frac{1}{2} m v^{2}$
$\Rightarrow v_{ma x} = \frac{F}{mk}$

mk​πF​

mk​2F​

mk​F​

πmk​F​

Maximum speed is at mean position (equilibrium). F=kx x=kF​ WF​+Wsp​=ΔKE F(x)−21​kx2=21​mv2−0 F(kF​)−21​k(kF​)2=21​mv2 ⇒vmax​=mk​F​

$\frac{π F}{mk}$

$\frac{2 F}{mk}$

$\frac{F}{mk}$

$\frac{F}{π mk}$

Maximum speed is at mean position (equilibrium). $F = k x$
$x = \frac{F}{k}$
$W_{F} + W_{s p} = Δ K E$
$F (x) - \frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} - 0$
$F (\frac{F}{k}) - \frac{1}{2} k (\frac{F}{k})^{2} = \frac{1}{2} m v^{2}$
$\Rightarrow v_{ma x} = \frac{F}{mk}$