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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Newton's Laws of Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Newton's Laws of Motion

A 2 kg block is suspended from the ceiling of an elevator using a...

A 2 kg2 \text{ kg} block is suspended from the ceiling of an elevator using a massless spring having a spring constant k=100 N/mk = 100 \text{ N/m}. The elevator is going up with an acceleration of 2.0 m/s22.0 \text{ m/s}^2. Find the elongation of the spring. If an additional 1 kg1 \text{ kg} is attached to the block, what would be the further elongation?

Options

  1. A0.20 m0.20 \text{ m} and 0.10 m0.10 \text{ m}
  2. B0.16 m0.16 \text{ m} and 0.08 m0.08 \text{ m}
  3. C0.24 m0.24 \text{ m} and 0.12 m0.12 \text{ m}
  4. D0.24 m0.24 \text{ m} and 0.36 m0.36 \text{ m}

Correct answer

C. 0.24 m0.24 \text{ m} and 0.12 m0.12 \text{ m}

Step-by-step solution

Given m1=2 kgm_1 = 2 \text{ kg}, k=100 N/mk = 100 \text{ N/m}, elevator upward acceleration a=2.0 m/s2a = 2.0 \text{ m/s}^2, and taking g=10 m/s2g = 10 \text{ m/s}^2.

Inside the accelerating elevator, the effective acceleration due to gravity is geff=g+a=10+2.0=12 m/s2g_{\text{eff}} = g + a = 10 + 2.0 = 12 \text{ m/s}^2.

At equilibrium, the restoring force balances the apparent weight of the block.

kx1=m1geffk x_1 = m_1 g_{\text{eff}}

100x1=2×12100 x_1 = 2 \times 12

x1=0.24 mx_1 = 0.24 \text{ m}

When an additional mass m2=1 kgm_2 = 1 \text{ kg} is attached, the additional restoring force balances the additional apparent weight.

kx2=m2geffk x_2 = m_2 g_{\text{eff}}

100x2=1×12100 x_2 = 1 \times 12

x2=0.12 mx_2 = 0.12 \text{ m}

The initial elongation is 0.24 m0.24 \text{ m} and the further elongation is 0.12 m0.12 \text{ m}.
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