- #1
decentfellow
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Homework Statement
Two blocks ##A## and ##B## having masses ##m_1= 1 kg, m_2 = 4 kg## are arranged as shown in the figure, The pulleys ##P## and ##Q## are light and frictionless. All the blocks are resting on a horizontal floor and the pulleys are held such that strings remain just taut.
At moment ##t=0## a force ##F = 30 t (N)## starts acting on the pulley P along vertically upward direction as shown in the figure. Calculate,
(i) the time when the blocks A and B loose contact with ground.
(ii) the velocity of A when B looses contact with ground
(iii) the height raised by A upto this instant.
(iv) the, work done by the force F upto this instant.
Homework Equations
$$F=30t \\ m_1=1 kg \\ m_2=4kg \\ F=ma$$
The Attempt at a Solution
(i)
FBD of pulley ##P##
From the above image it is clear that $$T=10t$$
FBDs of block ##A##, ##B## and the pulley ##Q##
From the FBD of pulley ##Q##, we get
$$T'=2T=20t$$
From the FBD of block ##A##, we get
$$N_1+T=m_1g$$
As the block ##A## is about to loose contact with the ground the normal reaction form the ground vanishes, i.e. ##N_1=0##
So, $$T=m_1g\implies 10t=1\times 10\implies t=1\text{ sec}$$
From the FBD of block ##B##, we get
$$T'+N_2=m_2g\implies 2T+N_2=m_2g$$
Same as for block ##A##, block ##B##, when about to lose contact with ground ##N_2=0##
$$\therefore 2T=m_2g\implies 20t=4\times 10\implies t=2\text{ sec}$$
(ii)
When ##B## loses contact with the ground, ##A## would have traveled with the acceleration ##a=10t## for ##t=1\text{ sec}##, i.e. from ##t=1 sec## to ##t=2 sec## so the velocity of the block ##A## as ##B## loses contact is found as follows:-
$$V=\int_{0}^{1}{10(t)}dt=5\text{ m/s}$$
(iii)When ##B## loses contact with the ground, ##A## would have traveled with the velocity ##a=10t## for ##t=1\text{ sec}##, i.e. from ##t=1 sec## to ##t=2 sec## so the displacement of the block ##A## from its initial position as ##B## loses contact is found as follows:-
$$X=\int{dx}=\int_{0}^{1}{vdt}=\int_{0}^{1}{(5t^2)dt}=\left(\dfrac{5}{3}t^3\right)^{1}_{0}=\dfrac{5}{3}m$$
(iv) Finally, the question where I am having trouble.
As the force acting on the system is ##F=30t##, but the force that pulls the block ##A## through a distance is the tension, so the work done should be $$\int_{0}^{1}{T.dx}=\int_{0}^{1}{(10t).(5t^2)dt}=\int_{0}^{1}{50t^3dt}=\dfrac{25}{2}J$$.
In my calculation of the work done by the force ##F##, I found it odd that I was using ##T## instead of ##F##, but according to what I know is that that the work done by a force acting on a body is given by the dot product of the force with the displacement vector, so, here the only massive body being displaced is the block ##A##, and the force acting on it is the tension, so I considered tension as the force that acts directly on the block. So, why is my answer different from that of the book's.