Question

A 2.30 kg block slides down a 16.0 m long ramp that makes an angle of 55.0degree with the horizontal. It starts at rest. (a) What is the block's speed at the bottom of the ramp? Use conservation of energy to find your answer. (b) The block collides and sticks to another block at rest that has 4 times its mass. What is the speed of the combined blocks? (c) A force stops the combined masses in a distance of 13.3 m. What is the magnitude of this force?

Answer 1

Here,

m = 2.30 Kg

L = 16 m

theta = 55 degree

a) let the speed at the bottom is v

0.5 * m * v^2 = m * g * L * sin(theta)

0.5 * v^2 = 9.8 * 16 * sin(55)

v = 16.03 m/s

the block's speed at the bottom of the ramp is 16.03 m/s

b) let the speed of combined mass is vf

Using conseravtion of momentum

m* v = (m + 4m) * vf

vf = 16.03/5 = 3.21 m/s

the final speed is 3.21 m/s

c)

let the magnitude of force is F

Using work energy theorum

0.5 * 5m * vf^2 = F * d

0.5 * 5 * 2.3 * 3.21^2 = F * 13.3

F = 4.45 N

the magnitude of force acting is 4.45 N