A vertical spring of force constant 100 Nm- 1 is attached to the ceiling, with a block of mass 10 kg suspended from it. Now an external force F is applied on the block so that the spring is stretched by an additional length of 2 m . The work done by the force F is [Take g=10 m s- 2 ]

Question

A vertical spring of force constant 100 Nm- 1 is attached to the ceiling, with a block of mass 10 kg suspended from it. Now an external force F is applied on the block so that the spring is stretched by an additional length of 2 m . The work done by the force F is [Take g=10 m s- 2 ] <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-rzxpx6odwfps.png" /> (A) 200J (B) 400J (C) 100J (D) 600J

Tardigrade Admin · Accepted Answer

x = elongation in spring due to mass 10 kg = (1 0 × 1 0/1 0 0) = text1 m mathrmW mathrmF=(1/2) × 100 ×[(3)2-(1)2]-10 × 10 × 2 = 200J

$200 J$

$400 J$

$100 J$

$600 J$

x = elongation in spring due to mass 10 kg
$= \frac{10 \times 10}{100} = 1 m$
$W_{F} = \frac{1}{2} \times 100 \times [(3)^{2} - (1)^{2}] - 10 \times 10 \times 2$
= $200 J$

Q. A vertical spring of force constant 100Nm−1 is attached to the ceiling, with a block of mass 10kg suspended from it. Now an external force F is applied on the block so that the spring is stretched by an additional length of 2m . The work done by the force F is [Take g=10ms−2 ]

200J

400J

100J

600J

x = elongation in spring due to mass 10 kg =10010×10​=1 m WF​=21​×100×[(3)2−(1)2]−10×10×2 = 200J

$200 J$

$400 J$

$100 J$

$600 J$

x = elongation in spring due to mass 10 kg
$= \frac{10 \times 10}{100} = 1 m$
$W_{F} = \frac{1}{2} \times 100 \times [(3)^{2} - (1)^{2}] - 10 \times 10 \times 2$
= $200 J$