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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 $0$?

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  • $\begingroup$ Isn't the work done by just the force F just the force multiplied by the distance the object travels? The problem says the force is always tangent to the trajectory. $\endgroup$ – Aaron Stevens Apr 19 '18 at 12:01
  • $\begingroup$ @AaronStevens consider the surface of the hill be some curved function f and the F is applied in the direction of the tangent line of the function f at that point, thats how I understood. But everything seems fine except why is the acceleration 0? $\endgroup$ – Nick Apr 19 '18 at 12:30
  • $\begingroup$ Probably because they say that the object is moved slowly. You can have a non-zero velocity with 0 acceleration. Since acceleration is 0, and F and friction act tangent to the trajectory, we know F = friction + tangent gravity force, and we know that the work done by F is proportional to the actual distance the object traveled. If the path is curvy then you would need to know the length of that curvy path. $\endgroup$ – Aaron Stevens Apr 19 '18 at 12:40
  • $\begingroup$ A word of advice : Instead of doing cumbersome integration, here, in this problem, the work-energy theorem comes very handy. Since the body is slowly hauled up the hill, the initial and final kinetic energy is zero, so the work done by individual forces, i.e, friction, weight, applied force summed together is zero. Then you can re-arrange to find the work done by the applied force. $\endgroup$ – vs_292 Apr 26 '18 at 15:34
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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.

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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.

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