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PhysicsNLMFBD & EquilibriumMedium2 minQB

PhysicsMediumnumerical

A 1 kg block B rests as shown on a bracket A of same mass. Constant forces $F_{1} = 20 N$ and $F_{2} = 8 N$ start to act at time $t = 0$ when the distance of block B from pulley is 50 cm . Time when block B reaches the pulley is

Answer:

0.50

Solution:

$a_{A} = \frac{20 - 16}{1} = 4 m / s^{2}$ $a_{B} = \frac{8}{1} = 8 m / s^{2}$

$a_{B / A} = 4 m / s^{2}$

$0.5 = 1/2 \times 4 t^{2}$

$t = \frac{1}{2} sec = 0.5 sec$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:FBD & Equilibrium

⏱ 2mℹ️ Source: QB

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