Question

A pendulum of mass m and length $\ell$ is suspended from the ceiling of a trolley which has a constant acceleration a in the horizontal direction as shown in figure. Work done by the tension is (In the frame of trolley) -

• A. $-mg\ell (1-cos\theta )$
• B. $ma\ell sin\theta$
• C. $ma\ell cos\theta$
• D. zero

Answer 1

The correct option is D zero

$\text{FBD of Pendulum}$

refer to image

So displacement in x direction $=-\left(\mathrm{lcos}\theta \right)$

$\begin{array}{cc}\text{and displacement in}y\text{direction}& =(\ell -l\sin \theta )\\ & =l-l\sin \theta \\ & =l(1-\sin \theta )\end{array}$

in upward direction

$\begin{array}{c}\text{So work done by gravity}\left(w_g\right)=-mgl(1-\sin \theta )\\ \text{and work done by Psuedo fore}\left(W_P\right)=\mathrm{mal}\cos \theta \\ \text{and we know}\tan \theta =g/a=\frac{mg}{ma}\end{array}$

$\begin{array}{c}\text{So By enagy conservation}\\ w_g+W_p+wt=△KE\\ \text{Kinetic Energy is (zero'at B).}\end{array}$

$\begin{array}{c}\text{So Put value of}K\epsilon \text{,}Wp\text{,}{W}_2\\ -\mathrm{mgl}(1-\sin \theta )+mal\cos \theta +WT=0\\ -mgl+mgl\sin \theta +mal\cos \theta +WT=0\end{array}$

$\begin{array}{c}\sin \theta =\frac{9}{\sqrt{9^2+a^2}}\text{and}\cos \theta =\frac{a}{\sqrt{g^2+a^2}}\\ \text{-mgl}+\text{mgl}\frac{g}{\sqrt{g^2+a^2}}+\text{mal}\frac{a}{\sqrt{g^2+a^2}}=\left(WT\right)\end{array}$

$WT=0$

$\begin{array}{c}\text{So work done by Tension is zero' ie WT}\\ \text{so Ans is (D).}\end{array}$