Question

A trolley $T$ (mass $5{\text{ }}kg$) on a horizontal smooth surface is pulled by a load ($2{\text{ }}kg$) through a uniform rope $ABC$ of length $2{\text{ }}m$ and mass $1{\text{ }}kg$. As the load falls from $BC{\text{ }} = {\text{ }}0$ to $BC{\text{ }} = {\text{ }}2{\text{ }}m$, its acceleration in $m{s^{-2}}$ changes from ?

Answer 1

Hint:We will consider the system for cases when the mass was not moving and when it is moving. Then, we will use the appropriate formula using Newton’s second law of motion for evaluating the initial and final accelerations.

Formula Used:
$F{\text{ }} = {\text{ }}m{\text{ }} \times {\text{ }}a$
$\Rightarrow a{\text{ }} = {\text{ }}\dfrac{F}{m}$

Complete step by step answer:
When the trolley, mass and the rope are at rest initially, then the force acting on the system will only be the weight of the mass. Thus,
$F{\text{ }} = {\text{ }}2{\text{ }} \times {\text{ }}g$
We take, acceleration due to gravity, $g{\text{ }} = {\text{ }}10{\text{ }}m{s^{-2}}$
Putting in the value, we get
$F{\text{ }} = {\text{ }}2{\text{ }} \times {\text{ }}10{\text{ }} = {\text{ }}20{\text{ }}N$
Now, total mass of the system will be the summation of the mass of the trolley, mass of the load and the mass of the rope.Thus,
$m{\text{ }} = {\text{ }}2{\text{ }} + {\text{ }}5{\text{ }} + {\text{ }}1{\text{ }} = {\text{ }}8{\text{ }}kg$
Thus, the initial acceleration will be,
$a{\text{ }} = {\text{ }}\dfrac{{20}}{8}{\text{ }}m{s^{-2}}$

Now, when the load moves from $0$ to $2{\text{ }}m$ and we know that the length of the string is also $2{\text{ }}m$. Thus, the new force will be constituted by the mass of the rope as well as the mass of the load.Thus, the new force will be
$F'{\text{ }} = {\text{ }}(2{\text{ }} + {\text{ }}1){\text{ }} \times {\text{ }}10{\text{ }} = {\text{ }}30{\text{ }}N$
Thus, the new acceleration will be,
$a'{\text{ }} = {\text{ }}\dfrac{{30}}{8}{\text{ }}m{s^{-2}}$
Hence, the acceleration changes from $\dfrac{{20}}{8}{\text{ }}m{s^{-2}}$ to $\dfrac{{30}}{8}{\text{ }}m{s^{-2}}$.
If we need to calculate the actual numerical value, then we get $2.5{\text{ }}m{s^{-2}}$ to $3.75{\text{ }}m{s^{-2}}$ .

Note:We considered the mass of the rope in the second case but not in the first case because in the first case only element under the influence of gravity will only be the load and not the rope whereas in the second case the rope moves $20{\text{ }}m$ and we are given that the length of the rope is also $20{\text{ }}m$. Hence, the total rope will also come under the influence of gravity and thus we considered the weight of the rope in the second case.