A 5kg block begins from rest and slides down on an inclined

A 5kg block begins from rest and slides down on an inclined plane. After 4s, the block has a velocity of 6m/s. If the angle of inclination of the plane is 45 degrees, approximately how far has the block travelled after 4s

A) 1.5 m

B) 3.0 m

C) 6.0 m

D) 12.0 m

Solution

From the equations of motions:

v = u + at

where v = velocity at time t,

u = velocity at time t = 0,

a = acceleration

We know, u = 0, v = 6 m/s at time t = 4 s,

Thus, 6 = 0 + a x 4

a = 1.5 m/s^2

Now, another equation of motion says that,

displacement, s = ut + 0.5 a t^2

s = (0 x 4) + (0.5 x 1.5 x 4²)

s = 12 m

Hence, And. (D) 12 m