Question

Consider the situation as shown in figure below. The acceleration of block of mass m is

& \text{A) }\dfrac{g}{3}\text{ up the plane} \\\ & \text{B) }\dfrac{g}{3}\text{ down the plane} \\\ & \text{C) }\dfrac{g}{2}\text{ up the plane} \\\ & \text{D) }\dfrac{g}{2}\text{ down the plane} \\\ \end{aligned}$$

Answer 1

We have given a diagram in which we have to find the acceleration for the block of mass m. Now there are two blocks of mass m and 2m respectively and there will be some force acting on it. So we will be redrawing the diagram and labelling the force on mass m and 2m. Then by balancing the forces we can find the acceleration of the block of mass m.

Complete answer:
Let us first redraw the given diagram to show forces acting on mass m and 2m.

If T is the tension in the string attached to the block of mass m, then the tension in the string attached to block 2m will be 2T. As we can see there are two components of T which is balancing the string attached to mass 2m.
Now if we consider block 2m and if it is moving with acceleration a downward, then the forces on it will be the downward gravitational pull 2mg, upward tension i.e. 2T and the force due to its motion which is 2ma. Neglecting the friction between the surfaces and considering the pulley and string massless, if we balance the forces we get

& 2mg-2T=2ma \\\ & \Rightarrow mg-T=ma\text{ }...............\text{(i)} \\\ \end{aligned}$$ Now according to the arrangement the motion of block m is due to the motion of block 2m, then the force due to its motion will be equal. If a’ is the acceleration of block of mass m then we can write $$\begin{aligned} & ma'=2m(a) \\\ & \Rightarrow a'=2a \\\ \end{aligned}$$ So the acceleration of mass m will be 2a (shown is also labelled in above diagram) Now the force which will be experienced by the block of mass m is the tension in the string T, N is the normal force and the downward gravitational pull mg. There will be a component of mg opposite to the tension. Also there will be force due to acceleration of the block of mass m, this force will be in direction of T. Again neglecting the friction and considering the string pulley attached to block as massless, balancing the forces, we will get $$\begin{aligned} & T-mg\sin 30{}^\circ =m(2a) \\\ & \Rightarrow T-mg\left( \dfrac{1}{2} \right)=2ma\text{ }\left( \because \sin 30{}^\circ =\dfrac{1}{2} \right) \\\ & \Rightarrow T-\dfrac{mg}{2}=2ma\text{ }..............\text{(ii)} \\\ \end{aligned}$$ Now adding equation (i) and (ii) we get $$\begin{aligned} & mg-T+T-\dfrac{mg}{2}=ma+2ma \\\ & \Rightarrow mg-\dfrac{mg}{2}=3ma \\\ & \Rightarrow \dfrac{mg}{2}=3ma \\\ & \Rightarrow a=\dfrac{mg}{2(3m)} \\\ & \Rightarrow a=\dfrac{g}{6} \\\ \end{aligned}$$ Now the acceleration of the block of mass m is 2a which gives the value of $$\dfrac{g}{3}$$. Now we have discussed above that the force due to its motion is in the direction of T that is in an upward direction. Therefore the acceleration will be up the plane. **So the correct answer is A.** **Note:** Here we had made assumptions like the surface is frictionless and the pulleys and string used are massless, if we consider all these quantities then the solution would get complicated to solve. Although by neglecting these factors we always get an approximate answer and here we got the right answer.