Aforce of 50n acts on a roller of mass 15 kg initially at rest on a frictionless surface. the roller travels 12 m while the force acts. calculate the work done on the roller

600 J

Explanation:

By definition, the work is the process due to which an applied force causes a mass to move, and can be calculated as the product of the component of the force applied along the direction of the movement, times the distance traveled.

In this case, is W= F*d= 50 N*12 m= 600 J

It can be showed, that the work done on the mass (in absence of friction ) is equal to the change in the kinetic  energy of the mass.

If the roller is initially at rest, this means the following:

ΔK= 1/2 m * vf²

Now, applying Newton's 2nd Law, F=m*a ⇒ a= F/m= 3.33 m/s²

If vo=0, we can write the following:

vf²=2*a*x= 2*3.33 m/s²*12m=79.98 m²/s²

Replacing in ΔK, we obtain

ΔK= 1/2*15 kg* 79.98 m²/s²≈ 600 J

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explanation:

it was a fall down when it happens

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