A 12.0-kg box is being pushed from the bottom to the top of a frictionless ramp. When the box is pushed at a constant velocity, the non-conservative pushing force does 58.0 J of work. How much work is done by the pushing force when the box starts from rest at the bottom and reaches the top of the same ramp with a speed of 1.50 m/s

work done = 71.5 J

Explanation:

Given that,

Energy = Fifty eight J

Mass of box = 12.0 Kg

Speed = 1.5 m/s

U = work done

The external force acts solely on the box and also the earth system. conjointly note that the work done by external force

= modification of energy of the system

U = work done - K

Also recall that the box is moving with constant speed therefore, there will not be modification in acceleration, hence, there'll be no modification in K.E..

Therefore, K is adequate to zero.

K= 0

U = work done = energy

U = 58.0 J

work done by pushing force = K + U

work done = final K.E. - initial K.E. + U 1

Initial K.E. = zero

Substitute the mandatory values into equation one

work done = 1/2m v² - zero + U

work done= 0.5 x 12.0 x 1.5² + 58.0

= 13.5+58.0

work done = 71.5 J

The world one by the pushing force is 71.5 J

the answer would be c