Question
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 ___________________.

0.5 s
Solution
To determine the time when block B reaches the pulley, we need to analyze the forces acting on block B and bracket A.
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Identify the forces: The force F2=8 N applied to the string creates a tension T=8 N in the string.
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Forces on block B: The tension T pulls block B to the left. Applying Newton's second law:
mBaB=T⟹1⋅aB=8⟹aB=8m/s2 (to the left).
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Forces on bracket A: Two horizontal forces act on bracket A:
- F1=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 2T and acts in the direction opposite to the applied force F2 relative to the pulley's motion. More simply, the force exerted by the string on the bracket A is 2T=2×8=16 N to the right.
Therefore, the net force on A is Fnet,A=F1 (left)−2T (right)=20−16=4 N (to the left). Applying Newton's second law:
mAaA=4⟹1⋅aA=4⟹aA=4m/s2 (to the left).
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Relative acceleration: Calculate the relative acceleration of block B with respect to bracket A. Let the left direction be positive.
aB/A=aB−aA=8m/s2−4m/s2=4m/s2 (to the left).
Since aB/A is to the left, block B moves towards the pulley.
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Kinematics: The initial distance of block B from the pulley is 50cm=0.5m. Using the kinematic equation for constant acceleration: s=v0t+21at2. Since the blocks start from rest, v0=0.
0.5=0+21(4)t2 0.5=2t2 t2=0.25 t=0.25=0.5s.
Therefore, the time when block B reaches the pulley is 0.5 s.