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The height of the hill is h and length of it's base is l.A force F which is always tangential is pulling the body up the hill. The friction coefficient between hill and body is $\mu$.The body is being pulled up slowly. The work done by force F is to be found.

I first used conservation of energy. Change in KE is 0 and change in PE is mgh. mgh is equal to the work done by F and friction. I am unable to figure out the work done by friction. Can you please help me out with this.

  • $\begingroup$ As written I do not think there is anybody who can find the work done by force $F$ against the frictional force. $\endgroup$ – Farcher Jan 26 at 17:17
  • $\begingroup$ @Farcher I tend to agree with you. But suppose it can be assumed that the contact area between the mass and the surface of the hill is constant (I think it would have to be otherwise the coefficient of kinetic friction would vary). Then given that the normal force on the hill will be greatest where the hill is flatter and less where the hill is steeper, couldn’t we assume the effective distance over which the friction force is constant is the projection $l$. Your thoughts. $\endgroup$ – Bob D Jan 26 at 18:11
  • $\begingroup$ @BobD Friction is a non-conservative force and so is path dependent. I this case the frictional force will vary from small at the start larger in the middle then smaller and then a maximum at the top so he frictional force is never constant. $\endgroup$ – Farcher Jan 26 at 22:46
  • $\begingroup$ @Farcher Agree, but isn't there an average friction force over the path? $\endgroup$ – Bob D Jan 26 at 22:49
  • $\begingroup$ @Farcher Let me put it this way. Consider the path to consist of vertical and horizontal components. For each vertical component the normal force and thus friction force is zero. For each horizontal component the normal force is mg and the friction force is umg. The sum of all the horizontal components of the path is simply the horizontal distance from the foot of the path to the end of the path, or $l$. $\endgroup$ – Bob D Jan 26 at 23:11

I tend to agree with @Farcher comment. You would have to make some assumptions to determine the friction work.

For one, you would have to assume the contact area between the mass and hill is constant as the mass is moved up the hill. I think it would have to be; otherwise the coefficient of friction would vary.

Now, if we can assume that the contact area is constant, we know the friction force is the normal force of the mass on the hill times the coefficient of friction and that that force opposes the applied force $F$. We can consider the path up the hill as composed of vertical and horizontal components.

For the vertical components of the path, the normal force and thus the friction force is zero. Thus the friction work done on the vertical components of the path is zero.

For the horizontal components of the path the normal force is $mg$ and the friction force is $μmg$.

Now ask yourself, what is the sum of all the horizontal components of the path. If you can answer this, you should have enough information to calculate the portion of the total work done by F as being friction work. (The rule on this site is not to provide answers to homework and exercises, but guidance). When you calculate the friction work, ask yourself where does this energy go?

Good luck and hope this helps.

  • $\begingroup$ Thanks @Bob D. I figured out the answer to be mgh + umgl and that turned out to be right. $\endgroup$ – Suven Jagtiani Jan 27 at 6:08
  • $\begingroup$ @SuvenJagtiani Are you sure about the sign of the friction work? Where does the friction work energy transfer go? $\endgroup$ – Bob D Jan 27 at 9:56

@BobD thank you for your useful comments which have shown my initial comment to be wrong.

Consider an object of mass $m$ moving from $A$ to $C$ either direct along the slope of incline $\theta$ to the horizontal or from $A$ to $B$ along the horizontal and then from $B$ to $C$ vertically upwards.

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Assuming that there is no acceleration of the object then the work done by the force $F$ along the second path is $\mu mg \Delta l + mg \Delta h$.

For the direct path from $A$ to $C$ along the slope is can be shown that $F=mg \sin \theta + \mu mg \cos \theta$.

So the work done by the force $F$ along this path is $F\, \dfrac {\Delta l}{\cos \theta}$ which works out to be $\mu mg \Delta l + mg \Delta h$ as before.

Now the path shown in the diagram can be thought of as the sum of many such incremental paths so the work done is $\mu mg l + mgh$.


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