Work done by moving an object up a hill

Question

I was doing this problem:

A body of mass m was slowly hauled up the hill by a force $F$ which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if height of the hill is $H$, the length of it's base is $L$ and the coefficient of the friction is $K$.

I started with summing up all forces that acts on the object

$$\sum{F_{total} = F -mg\sin\theta - Kmg\cos\theta = F - mg(\sin\theta +K\cos\theta) = ma}$$

then

$$dW = \sum{F_{total} \cdot \mathrm ds =[F - mg(sin\theta +K\cos\theta)}]\cdot\mathrm ds \tag{**}$$

then I assumed the height of the hill depends on the length of the base, so I wrote $f(x)=h$, $ $ $f(L)=H $ and $f(0) = 0$

then I wrote $$ \cos\theta =\frac{1}{\sqrt{1+(f^{'}(x))^2}},\qquad \sin\theta = \frac{f^{'}(x)}{\sqrt{1+(f^{'}(x))^2}}, \qquad ds = \sqrt{1+(f^{'}(x))^2}\,\mathrm dx$$

I plugged those into the $(**)$ equation and got this: $$ dW=\big[F-mg(f^{'}(x) +K)\big]dx \\ W=\int_0^{L}\big[F-mg(f^{'}(x) +K)\big]dx \\ W=\int_0^{L}Fdx - mg\int_0^{L}(f^{'}(x)+K)dx\\ W=\int_0^{L}Fdx -mg(H+KL) $$ So I am stuck here trying to find what $F$ is...

Edit:

I think I did it, so here it is :

$$ W=\int_0^{L}\sum{F}=\int_0^{L}F\,\mathrm dx -mg(H+KL)\\ \int_0^{L}ma\,\mathrm dx=\int_0^{L}F\,\mathrm dx -mg(H+KL)\\ W_F=\int_0^{L}F\,\mathrm dx=maL +mg(H+KL)\\$$

Answer: $W_F=maL +mg(H+KL)$

I guess this is for general case because in my textbook answer was $W_F=mg(H+KL)$ so, I suspect $ a = 0$ in this case, but why acceleration is $0$?

Answer 1

The applied force does work on the body. This work can achieve three things here : (i) it can overcome friction ($mgKL$), (ii) it can increase the gravitational potential energy of the body ($mgH$), and (iii) it can increase the kinetic energy of the body ($maL$). The problem states that the body is "slowly hauled" which implies that there is no increase in its kinetic energy, which means there is no acceleration ($a=0$). The force varies so as to ensure that there is no acceleration.

If the applied force is constant there may in fact be acceleration on some parts of a non-linear trajectory, but then there is deceleration on others. The body ends with the same slow speed (and kinetic energy) as it started, so there is no acceleration.

Answer 2

Whenever the question says “slowly” it generally means that the force is just enough to cancel other forces and just get the object in motion with a uniform velocity and no acceleration.