Question

The coefficient of friction between the block of mass m₁ = 3.00 kg and the surface in the figure below is μ_k = 0.455. The system starts from rest. What is the speed of the ball of mass m₂ = 5.00 kg when it has fallen a distance h = 1.10 m?

Answer 1

from free body diagram of an above figure, we get

a = (m₂ g - _k m₁ g) / (m₁ + m₂)                                                                 { eq.1 }

where, m₁ = mass of the first block = 3 kg

m₂ = mass of the second block = 5 kg

_k = coefficient of kinetic friction between the block & surface = 0.455

g = acceleration due to gravity = 9.8 m/s²

inserting these values in above eq.

a = [(5 kg) (9.8 m/s²) - (0.455) (3 kg) (9.8 m/s²)] / [(3 kg) + (5 kg)]

a = (35.7 kg.m/s²) / (8 kg)

a = 4.46 m/s²

using equation of motion 2,

h = v_o t + (1/2) a t²                                                                   { eq.2 }

where, h = fallen height = 1.1 m

v₀ = initial speed = 0 m/s

inserting the values in eq.2.

(1.1 m) = (0 m/s) t + (0.5) (4.46 m/s²) t²

t² = (1.1 m) / (2.23 m/s²)

t = 0.493 s²

t = 0.702 sec

Now, final speed of the ball which will be given as :

using equation of motion 1,

v = v₀ + at                                                                   { eq.3 }

inserting the values in eq.3,

v = (0 m/s) + (4.46 m/s²) (0.702 s)

v = 3.13 m/s