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Qns On Work and Energy

  • Thread starter Delzac
  • Start date
389
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hi, here's the Qns,

A 100-lb block of ice slides down an incline 5.0 ft and 3.0 ft high. A man pushes up on the ice parallel to the incline so that is slides down at constant speed. The coefficient of friction between the ice and the incline is 0.10. Find
(a) the force exerted by the man,
(b) the work done by the man on the block,
(c) the work done by gravity on the block,
(d) the work done by the surface of the incline on the block,
(e) the work done by the resultant force on the block, and
(f) the change in kinetic energy of the block.

I can't do Qns (a), (b)(i shall leave the rest alone for now)

What i did was to calculate frictional force 1st,
[tex]F_f = (0.1)(100) = 10 ft.lb[/tex] correct?

and then, i got stuck........ the Qns did state the angle at which the incline is tilted, so i was unable to resolve the forces(weight of ice).

Any help will be greatly appreciated. :)
 

Hootenanny

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Delzac said:
What i did was to calculate frictional force 1st,
[tex]F_f = (0.1)(100) = 10 ft.lb[/tex] correct?
I'm afraid not. Have you drawn a diagram? If not I would recommend doing so. Remember that the normal force always acts perpendicular to the surface. Now is gravity acting perpendicular to the surface? If not you will have to resolve it to find the component that is. Does that make sense?
 
389
0
ohh i see! the angle at which it is inclined is [tex]\sin^-1 (\frac{3}{5})[/tex] doing so allow us to resolve the forces, thx! i got it.
 
Last edited:

Hootenanny

Staff Emeritus
Science Advisor
Gold Member
9,598
6
Delzac said:
ohh i see! the angle at which it is inclined is [tex]\sin^-1 \frac{3}{5}[/tex] doing so allow us to resolve the forces, thx! i got it.
No problem, remember for the work questions that the work done is the product of the force and the distance moved in the direction of that force.
 

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