How much work was done to overcome friction?

  • Thread starter Lil_Aziz1
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  • #1
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Homework Statement


NO CALCULATOR ALLOWED.

a 50-kilogram box is pushed up a 1.5m-meter incline with an effort of 200 newtons. The top of the incline is 0.5 meter above the ground. How much work was done to overcome friction?

(A) 55J
(B) 75J
(C) 245J
(D) 300J
(E) 490J

The inclination in the question above is an example of a simple machine. On the basis of the information provided, what is the approximate efficiency of this machine?


(A) 50%
(B) 100%
(C) 20%
(D) 10%
(E) 80%

Homework Equations



[tex]W = Fd [/tex]

[tex]F_{fric} = \mu F_{n} [/tex]

[tex]W = E [/tex]

[tex] e = \frac{\Delta W}{W_{tot}} [/tex]

The Attempt at a Solution



Alright so the total work has to be 200*1.5; thus, [tex] W_{TOT} = 200*1.5 = W_{fric} + W_{frictionless} [/tex]

Or

[tex]W_{nc} = W_{TOT} - W_{frictionless} [/tex]

and idk what to do from here ;\

the efficiency part:

e =
 
Last edited:

Answers and Replies

  • #2
258
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Well how much work would it be if there was no friction? Or in other words, how much energy would it take to push the box up the ramp if the only force your doing work against is gravity?
 
Last edited:
  • #3
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Force * distance. right?
 
  • #4
258
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Yes but F Is different if there is no friction. Think about gravity being a conservative force and work is independent of the path taken.
 
  • #5
21
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ok so the 200N is the force with friction. how am i suppose to find the force without friction? Isn't the force without friction greater than 200 if you think about it intuitively? If friction is not acting on the box, the only force acting on the box is the x-component of gravity; thus, [tex]F = F_{g}\sin ((((((((\arctan (\frac{0.5}{1.5}) [/tex]. But, [tex]\arctan (\frac{1}{3}) [/tex] is not given, and we can't use a calculator.
 
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  • #6
258
1
How much work does it take to lift a 1kg box 3m high? W=mgh=(1kg)(9.8m/s^2)(3m)
 
  • #7
21
1
oh! so to get the work done by friction, one would subtract 200*1.5 from (10)(3). Thanks a ton!

To get the efficiency, one would just do (mgh)/(F*d)
 

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