Question

$F$ acts always such that it is tangential to the surface of hill
Find the work done by force $F$

• A. $mgl$
• B. $mg\sqrt{{\ell }^2+h^2}$
• C. $\mu mgl+mgh$
• D. $\mu mgh+mgl$

Answer 1

The correct option is C $\mu mgl+mgh$

Ref. image 1

According to work energy theorem we know

$(W_{A-B}{)}_{net}=K{E}_B-K{E}_A$

Hence $(W_{A-B}{)}_{net}=$ work done from A to B (net) and KE = kinetic energy

as $V_B=V_A,K{E}_B=K{E}_A$

$\therefore (W_{A-B}{)}_{net}=0$

Now from acting on mass m are.

By FBD

Ref. image 2

So work done by all the forces

$W_{net}=D$

Work done by $F+$ work done by $mg+$ work done by friction = 0

$W_F=-W_{mg}-W_{friction}$

$W_f=-(-mgh)-W_{friction}$ taking upwind direction

$=mgh-W_{friction}$

Now for $W_{friction}$

Now for $W_{friction}$

ref. image 3

friction force = $\mu mg\cos \theta$ as $(N=mg\cos \theta )$

Let the mass more ds.

ref. image 4 Then $d{W}_{friction}=-mumg\cos \theta$

Ref. image 5

$W_{friction}=-\mu mg\int ds\cos \theta$

$=-\mu mg\int dx$

$=-\mu mgl$

$\therefore W_f=mgh+\mu mgl$

$=mg(h+\mu l)$