Question

A mass m1 on a horizontal shelf is attached by a thin string that passes over a frictionless pulley to a 2.4 kg mass (m2) that hangs over the side of the shelf 1.2 m above the ground. The system is released from rest at t = 0 and the 2.4 kg mass strikes the ground at t = 0.86 s. The system is now placed in its initial position and a 1.2 kg mass is placed on top of the block of mass m1. Released from rest, the 2.4 kg mass now strikes the ground in 1.3 seconds.

(a) Determine the mass m₁.

(b) Determine the coefficient of kinetic friction between m₁ and the shelf.

Answer 1

Case 1:

distance travelled = d = 1.2 m

time taken = t = 0.86 sec

initial velocity = V_o = 0 m/s

acceleration = a

using the equation

d = V_o t + (0.5) a t²

1.2 = 0 x 0.86 + (0.5) a (0.86)²

a = 3.25 m/s²

force equation for hanging mass m₂ is given as

m₂ g - T = m₂ a

2.4 x 9.8 - T = 2.4 x 3.25

T = 15.72 N

force equation for mass m₁ on the shelf is given as

T - f = m₁ a

15.72 - m₁(9.8) = 3.25 m₁

m₁ = 15.72 / ((9.8) + 3.25)                        eq-1

Case 2 :

distance travelled = d = 1.2 m

time taken = t = 1.3 sec

initial velocity = V_o = 0 m/s

acceleration = a

using the equation

d = V_o t + (0.5) a t²

1.2 = 0 x 1.3 + (0.5) a (1.3)²

a = 1.42 m/s²

force equation for hanging mass m₂ is given as

m₂ g - T = m₂ a

2.4 x 9.8 - T = 2.4 x 1.42

T = 20.11 N

force equation for mass m₁+ 1.2 on the shelf is given as

T - f = (m₁+ 1.2) a

20.11 - (m₁+ 1.2)(9.8) = 1.42 (m₁+ 1.2)

(m₁+ 1.2) = 20.11 / ((9.8) + 1.42)                        eq-2

15.72 / ((9.8) + 3.25) + 1.2 = 20.11 / ((9.8) + 1.42)  

= 0.52

using eq-1

m₁ = 15.72 / ((9.8) + 3.25)  

m₁ = 15.72 / ((0.52)(9.8) + 3.25)

m₁ = 1.88 kg