A box of mass 2 kg has an initial speed of 10 m/s at the foot of the ramp. Given that the friction along the ramp is 2 N, calculate the height h that it reaches when the speed of the box is 5 m/s. (g = 10 m/s^{2 }) ** **

**Solutions:**

For this question, the additional thing to note is the work done against friction. All the energy possesses by the box at **bottom is KE.** This KE will decrease and be converted to **remaining KE at height h + gain in GPE + work done against friction**.

To find the work done against friction, we need to find the distance moved by the box on the inclined ramp, d.

sin60 = h / d

d = h / sin60

Conservation of energy,

KE at bottom = KE at h + GPE at h + friction force

1/2mv^{2} = 1/2mv^{2} + mgh + F x d

(1/2 x 2 x 10^{2}) = (1/2 x 2 x 5^{2}) + (2 x 10 x h) + (2 x h/sin60)

h = 3.36 m