Question

A 20 kg block starting from rest slides up a 300 inclined plane under the action of a 175 N force directed

along the inclined plane. The coefficient of kinetic friction between the block and the plane is 0.2.

Determine the (i) speed of the block after it slides 4.5 m and (ii) the distance travelled by the block

when its speed becomes4.5 m/s.

Answer 1

Answer:

F = ma = 175 - mg sin 30 - u mg cos 30

F = 175 - 20 * 9.8 * (.5 + .2 * .866) = 43.05 N

a = F / m) = 43.05 / 20  = 2.15 m/s^2

2 a s = v^2   since the initial speed is zero

v = (2 * 2.15 * 4.5)^1/2 = 4.40 m/s

s = v^2 / (2 * a) = 4.5^2 / (2 * 2.15) = 4.71 m