Question

A block of mass $m$ is stationary with respect to a rough wedge as shown in figure. Starting from rest in time $t$.
(Given that $m=1\text{kg},\theta ={30}^{\circ },a=2{\text{m/s}}^2,t=4\text{s}$)
Match Column $I$ with column $II$
$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{(a) Work done on block by gravity}& \text{(p)}144\text{J}\\ \text{(b) Work done on block by normal reaction}& \text{(q)}32\text{J}\\ \text{(c) Work done on block by friction}& \text{(r)}56\text{J}\\ \text{(d) Work done on block by all the forces}& \text{(s)}48\text{J}\\ & \text{(t)}\text{None}\end{array}$

• A. $a-t,b-p,c-s,d-q$
• B. $a-p,b-t,c-s,d-q$
• C. $a-p,b-t,c-q,d-q$
• D. $a-q,b-t,c-s,d-p$

Answer 1

The correct option is A $a-t,b-p,c-s,d-q$

We know that as body starts from rest,
$S=\frac{1}{2}a{t}^2=\frac{1}{2}\left(2\right)4^2=16\text{m}v=at=8\text{m/s}$


From Newton's second law,
$N=m(g+a)\cos {30}^{\circ }f=m(g+a)\sin {30}^{\circ }$

$W_{mg}=mgS\cos {180}^{\circ }=10\times 16\times (-1)=-160\text{J}$
$W_N=NS\cos {30}^{\circ }=m(g+a)S{\cos }^2{30}^{\circ }=12\times 16\times \frac{3}{4}=144\text{J}$
$W_f=fS\cos {60}^{\circ }=m(g+a)S\sin {30}^{\circ }\cos {60}^{\circ }=12\times 16\times \frac{1}{4}=48\text{J}$
$W_{NetF}=\Delta KE=\frac{1}{2}m{v}^2=\frac{1}{2}(at{)}^2=32\text{J}$