A 20 kg box has an initial velocity of 2 m/s starting at the bottom of a 30-degree inclined plane. A person pushes on the box directly up the frictionless inclined plane so that it travels up the inclined plane at a constant velocity of 2 m/s. Calculate the how much is done by the person after 5 seconds have past.

Explanation:

The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .

The weight component acting on box parallel to incline plane

= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N

This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .

Force exerted by person = 98 N

distance travelled in 5 s

= velocity x time

= 2 x 5 = 10 m

Work done by person

= 98 x 10

= 980 J .

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

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