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mina [271]
3 years ago
10

You push your couch a distance of 3.9 m across the living room floor with a horizontal force of 220.0 n. the force of friction i

s 144.0 n. what is the work done by you, by the friction force, by gravity, and by the net force?
Physics
1 answer:
Mrac [35]3 years ago
4 0

The general formula to calculate the work is:

W=Fd \cos \theta

where F is the force, d is the displacement of the couch, and \theta is the angle between the direction of the force and the displacement. Let's apply this formula to the different parts of the problem.


(a) Work done by you: in this case, the force applied is parallel to the displacement of the couch, so \theta=0^{\circ} and \cos \theta=1, therefore the work is just equal to the product between the horizontal force you apply to push the couch and the distance the couch has been moved:

W=Fd=(220.0 N)(3.9 m)=858 J


(b) work done by the frictional force: the frictional force has opposite direction to the displacement, therefore \theta=180^{\circ} and \cos \theta=-1. Therefore, we must include a negative sign when we calculate the work done by the frictional force:

W=-Fd=-(144.0 N)(3.9 m)=-561.6 J


(c) The work done by gravity is zero. In fact, gravity (which points downwards) is perpendicular to the displacement of the couch (which is horizontal), therefore \theta=90^{\circ} and \cos \theta=0: this means

W=0.


(d) Work done by the net force:

The net force is the difference between the horizontal force applied by you and the frictional force:

F=220 N-144 N=76 N

And the net force is in the same direction of the displacement, so \theta=0^{\circ} and \cos \theta=1 and the work done is

W=Fd=(76 N)(3.9 m)=296.4 J


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Answer:

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Explanation:

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                       v = u + at  

                       16.67 = 0 + a x 5.4

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b) We have equation of motion s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Acceleration, a = 3.09 m/s²

         Time, t = 5.4 s      

      Substituting

                       s = ut + 0.5 at²

                       s = 0 x 5.4 + 0.5 x 3.09 x 5.4²

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c) We have equation of motion s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Acceleration, a = 3.09 m/s²

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      Substituting

                       s = ut + 0.5 at²

                       250 = 0 x t + 0.5 x 3.09 xt²

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