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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Physics
Chapter
Work, Power and Energy

BITSAT 2024 · Physics · Work, Power and Energy

An ideal massless spring S can be compressed 1 m by a force of 100...

An ideal massless spring SS can be compressed 1 m by a force of 100 N in equilibrium. The same spring is placed at the bottom of a frictionless inclined plane inclined at 30∘30^{{}^{\circ}} to the horizontal. A 10 kg block MM is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring by 2 m. If g=10m/s2g=10 \mathrm{ m} / \mathrm{s}^2, what is the speed of mass just before it touches the spring? An ideal massless spring S can be compressed 1 m by a force of 100 N in equilibrium. The same spring is placed at the

Options

  1. A20m/s\sqrt{20} \mathrm{ m} / \mathrm{s}
  2. B30m/s\sqrt{30} \mathrm{ m} / \mathrm{s}
  3. C10m/s\sqrt{10} \mathrm{ m} / \mathrm{s}
  4. D40m/s\sqrt{40} \mathrm{ m} / \mathrm{s}

Correct answer

A. 20m/s\sqrt{20} \mathrm{ m} / \mathrm{s}

Step-by-step solution

F=kx∴k=Fx=1001N/m=100N/mF=k x\therefore \quad k=\frac{F}{x}=\frac{100}{1} \mathrm{ N} / \mathrm{m}=100 \mathrm{ N} / \mathrm{m} Now, from energy conservation, between natural length of spring and its maximum compression state. 12mv2+mgh=12kxmax⁡2\frac{1}{2} m v^2+m g h=\frac{1}{2} k x_{\max }^2∴v=kxmax⁡2m−2gh=(100)(2)210−(2)(10)(22)=20m/s\begin{aligned} \therefore \quad v & =\sqrt{\frac{k x_{\max }^2}{m}-2 g h} \\ & =\sqrt{\frac{(100)(2)^2}{10}-(2)(10)\left(\frac{2}{2}\right)}=\sqrt{20} \mathrm{ m} / \mathrm{s}\end{aligned}$$
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