Question

Correct Answer is given. Solve and Explain

Answer 1

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\line(1,0){55}}\put(22.5,0){\framebox(10,10){\sf{m}}}\put(32.5,0){\vector(1,0){10}}\put(22.5,5){\vector(-1,0){10}}\put(22.5,5){\vector(-1,1){10}}\multiput(22.5,5)(5,5){2}{\vector(0,1){10}}\put(27.5,0){\vector(0,-1){10}}\put(1.5,4){ \sf{F\cos\theta} }\put(15,16){ \sf{F\sin\theta} }\put(9.5,12){\sf{F}}\put(17,6){ \theta }\put(29,19){\sf{R}}\put(25,-13){\sf{mg}}\put(37.5,1.5){\sf{f}}\end{picture}$

Since the net vertical force acting on the block is zero,

$\longrightarrow\sf{R+F\sin\theta=mg}$

$\longrightarrow\sf{R=mg-F\sin\theta}$

Since the block moves over the surface with constant velocity, net horizontal acceleration of the block is zero, so is net horizontal force.

$\longrightarrow\sf{F\cos\theta=f}$

$\longrightarrow\sf{F\cos\theta=\mu\,R}$

$\longrightarrow\sf{F\cos\theta=\mu(mg-F\sin\theta)}$

$\longrightarrow\sf{F\cos\theta=\mu\,mg-\mu\,F\sin\theta}$

$\longrightarrow\sf{F\cos\theta+\mu\,F\sin\theta=\mu\,mg}$

$\longrightarrow\sf{F(\cos\theta+\mu\,\sin\theta)=\mu\,mg}$

$\longrightarrow\sf{F=\dfrac{\mu\,mg}{\cos\theta+\mu\,\sin\theta}}$

Hence the work done by the applied force F is,

$\longrightarrow\sf{W=Fd\cos\theta}$

$\longrightarrow\sf{\underline{\underline{W=\dfrac{\mu\,mgd\cos\theta}{\cos\theta+\mu\,\sin\theta}}}}$

Hence (2) is the answer.