Question

A 4 kg block is launched up a 30° ramp with an initial speed of 5 m/s. The coefficient of kinetic friction between the block and the ramp is 0.25. What is the speed of the block (in m/s) when it has returned to the bottom of the ramp?

Answer 1

Answer:

The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.

Explanation:

Given;

mass of block, m =  4 kg

coefficient of kinetic friction, μk = 0.25

angle of inclination, θ = 30°

initial speed of the block, u = 5 m/s

From Newton's second law of motion;

F = ma

a = F/m

Net horizontal force;

∑F = mgsinθ + μkmgcosθ

$a = \frac{F_{NET}}{m} = \frac{mgsin \theta + \mu_kmgcos \theta}{m} \\\\a = gsin \theta + \mu_kgcos \theta\\\\a = 9.8sin (30) + 0.25*9.8cos(30)\\\\a = 4.9 + 2.1217\\\\a = 7.022 \ m/s^2$

At the  top of the ramp, energy is conserved;

Kinetic energy = potential energy

¹/₂mv² = mgh

¹/₂ v² = gh

¹/₂ x 5² = 9.8h

12.5 = 9.8h

h = 12.5/9.8

h = 1.28 m

Height of the ramp is 1.28 m

Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;

v² = u² + 2ah

v²  = 5² + 2 x 7.022 x 1.28

v²  = 25 + 17.976

v²  = 42.976

v = √42.976

v = 6.56 m/s

Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.