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Question: A 1kg block B rests as shown on a smooth bracket A of same mass. Constant forces F₁ = 20N and F₂ = 8...

A 1kg block B rests as shown on a smooth bracket A of same mass. Constant forces F₁ = 20N and F₂ = 8N start to act at time t = 0 when the distance of block B from pulley is 50cm. Time when block B reaches the pulley is ___________________.

Answer

0.5 s

Explanation

Solution

To determine the time when block B reaches the pulley, we need to analyze the forces acting on block B and bracket A.

  1. Identify the forces: The force F2=8F_2 = 8 N applied to the string creates a tension T=8T = 8 N in the string.

  2. Forces on block B: The tension TT pulls block B to the left. Applying Newton's second law:

    mBaB=T    1aB=8    aB=8m/s2m_B a_B = T \implies 1 \cdot a_B = 8 \implies a_B = 8 \, \text{m/s}^2 (to the left).

  3. Forces on bracket A: Two horizontal forces act on bracket A:

    • F1=20F_1 = 20 N acts to the left.
    • The string passing over the pulley exerts a force on the pulley (which is part of A). This force is 2T2T and acts in the direction opposite to the applied force F2F_2 relative to the pulley's motion. More simply, the force exerted by the string on the bracket A is 2T=2×8=162T = 2 \times 8 = 16 N to the right.

    Therefore, the net force on A is Fnet,A=F1 (left)2T (right)=2016=4F_{\text{net}, A} = F_1 \text{ (left)} - 2T \text{ (right)} = 20 - 16 = 4 N (to the left). Applying Newton's second law:

    mAaA=4    1aA=4    aA=4m/s2m_A a_A = 4 \implies 1 \cdot a_A = 4 \implies a_A = 4 \, \text{m/s}^2 (to the left).

  4. Relative acceleration: Calculate the relative acceleration of block B with respect to bracket A. Let the left direction be positive.

    aB/A=aBaA=8m/s24m/s2=4m/s2a_{B/A} = a_B - a_A = 8 \, \text{m/s}^2 - 4 \, \text{m/s}^2 = 4 \, \text{m/s}^2 (to the left).

    Since aB/Aa_{B/A} is to the left, block B moves towards the pulley.

  5. Kinematics: The initial distance of block B from the pulley is 50cm=0.5m50 \, \text{cm} = 0.5 \, \text{m}. Using the kinematic equation for constant acceleration: s=v0t+12at2s = v_0 t + \frac{1}{2} a t^2. Since the blocks start from rest, v0=0v_0 = 0.

    0.5=0+12(4)t20.5 = 0 + \frac{1}{2} (4) t^2 0.5=2t20.5 = 2 t^2 t2=0.25t^2 = 0.25 t=0.25=0.5st = \sqrt{0.25} = 0.5 \, \text{s}.

Therefore, the time when block B reaches the pulley is 0.5 s.