Two blocks A and B of masses 1 kg and 2 kg respectively are connected by a string, passing over a light frictionless pulley. Both the blocks are resting on a horizontal floor and the pulley is held such that string remains just taut.At moment t = 0, a force F = 20 t newton starts acting on the

Question

Two blocks A and B of masses 1 kg and 2 kg respectively are connected by a string, passing over a light frictionless pulley. Both the blocks are resting on a horizontal floor and the pulley is held such that string remains just taut.At moment t = 0, a force F = 20 t newton starts acting on the pulley along vertically upward direction as shown in figure. Calculate
(a) velocity of A when B loses contact with the floor.
(b) height raised by the pulley upto that instant.
(Take, g = 10 m/s²)

Accepted Answer

(a) Let T be the tension in the string. Then

2 T=20 t

or [latex]T=10 t ext { newton }[/latex]

Let the block A loses its contact with the floor at time [latex]t=t_{1}[/latex] . This happens when the tension in string becomes equal to the weight of A. Thus,

T = mg

or [latex]10 t_1=1 imes 10[/latex] ...(i)

or [latex]t_1=1 \mathrm{~s}[/latex]

Similarly, for block B, we have

[latex]10 t_2=2 imes 10[/latex]

or [latex]t_2=2 \mathrm{~s}[/latex] ...(ii)

i.e. the block B loses contact at 2 s. For block A, at time t such that t ≥ [latex]t_{1}[/latex] let a be its acceleration in upward direction. Then,

[latex]10 t-1 imes 10=1 imes a=(d v / d t)[/latex]

or [latex]d v=10(t-1) d t[/latex] ...(iii)

Integrating this expression, we get

[latex]\int_0^v d v=10 \int_1^t(t-1) d t[/latex]

or [latex]v=5 t^2-10 t+5[/latex] ...(iv)

Substituting t = [latex]t_{2}[/latex] = 2 s

[latex]v=20-20+5=5 \mathrm{~m} / \mathrm{s}[/latex] ...(v)

(b) From Eq. (iv), [latex]d y=\left(5 t^2-10 t+5 ight) d t[/latex] ...(vi)

where, y is the vertical displacement of block A at time [latex]t\left(\geq t_1 ight)[/latex]
Integrating, we have

[latex]\begin{aligned} \int_{y=0}^{y=h} d y &=\int_{t=1}^{t=2}\left(5 t^2-10 t+5 ight) d t \ h &=5\left[\frac{t^3}{3} ight]_1^2-10\left[\frac{t^2}{2} ight]_1^2+5[t]_1^2=\frac{5}{3} \mathrm{~m} \end{aligned}[/latex]

∴ Height raised by pulley upto that instant [latex]=\frac{h}{2}=\frac{5}{6} \mathrm{~m}[/latex]

Question 8.SE.21: Two blocks A and B of masses 1 kg and 2 kg respectively are ...

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