A vertical spring of force constant 100 Nm- 1 is attached with a hanging mass of 10kg . Now an external force is applied on the mass so that the spring is stretched by an additional 2m . The work done by the external force is: ( g=10 ms- 2 )

Question

A vertical spring of force constant 100 Nm- 1 is attached with a hanging mass of 10kg . Now an external force is applied on the mass so that the spring is stretched by an additional 2m . The work done by the external force is: ( g=10 ms- 2 ) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-zurvz7qepshj.png" /> (A) 200 J (B) 400 J (C) 450 J (D) 600 J

Tardigrade Admin · Accepted Answer

At equilibrium, mg=k x0 ⇒ x0=(m g/k)=1 m <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-zurvz7qekdqg.png" /> ∴ W=U2 - U1 =(1/2) kx22- [(1/2) k x12 + m g h] =(1/2) k (x22 - x12) - mgh =(1/2) × 100 × (32 - 12) - 10 × 10 × 2=200 J

Q. A vertical spring of force constant $100 N m^{- 1}$ is attached with a hanging mass of $10 k g$ . Now an external force is applied on the mass so that the spring is stretched by an additional $2 m$ . The work done by the external force is : ( $g = 10 m s^{- 2}$ )

$200 J$

$400 J$

$450 J$

$600 J$

At equilibrium, $m g = k x_{0} \Rightarrow x_{0} = \frac{m g}{k} = 1 m$

$∴ W = U_{2} - U_{1}$
$= \frac{1}{2} k x_{2}^{2} - [\frac{1}{2} k x_{1}^{2} + m g h]$
$= \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) - m g h$
$= \frac{1}{2} \times 100 \times (3^{2} - 1^{2}) - 10 \times 10 \times 2 = 200 J$

Q. A vertical spring of force constant 100Nm−1 is attached with a hanging mass of 10kg . Now an external force is applied on the mass so that the spring is stretched by an additional 2m . The work done by the external force is : ( g=10ms−2 )

200J

400J

450J

600J