a 2.0 kg object is pushed up from rest on an incline by a horizontal force F=15N

Question

a 2.0 kg object is pushed up from rest on an incline by a horizontal force F=15N. There is no friction between the object and incline. Between the incline from the floor is 30 degrees. (When the object has moved 2.5 m along the incline, I found the work done by the normal force on the object to be 0J, the work done by gravity to be -42.4J, the work done by the additional force F to be 32.5 J. I also found the kinetic energy of the object to be 4.95J and the speed to be 4.95m/s.)

The force F ceases to be applied at this location and as a result the object eventually slides down the incline. Find the kinetic energy and speed of the object as it returns to the bottom level of the incline.

Explanation / Answer

we have,

u = 4.95 m/s
v= 0
a = -gsin(30) = -4.9 m/s^2

v^2 = u^2 +2as
Distance = s = 2.5 meter
Total distance s'= 2.5 + 2.5 = 5 m
Height = h = s 'x sin(30) = 2.5 meter

a.

Kinetic ENergy 0.5mv^2 = mgh = 2 x 9.8 x 2.5 = 49 Joules

b.

0.5mv^2 = 49

0.5 x 2 v^2 = 49

SPeed = v = 7 m/s

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