A mass m1 on a horizontal shelf is attached by a thin string that passes over a

Question

A mass m1 on a horizontal shelf is attached by a thin string that passes over a frictionless pulley to a 2.5 kg mass (m2) that hangs over the side of the shelf 1.5 m above the ground. The system is released from rest at t = 0 and the 2.5 kg mass strikes the ground at t = 0.71 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 2.5 kg mass now strikes the ground 1.3 seconds later. (a) Determine the mass m1. (b) Determine the coefficient of kinetic friction between m1 and the shelf.

Explanation / Answer

For mass m2

m2g - T = m2a1 -------------(1)

For mass m1

T - f = m1a1         -------------(2)

The acceleration of the masses

S =0.5 a1 t2

a1 = 2S / t2

= 2 * 1.5 / 0.71*0.71

= 5.95 m/s2

Tension in the string is

T =m2g - m2a

= 2.5 * 9.8 - 2.5 * 5.95 = 9.63 N

When mass 1.2 kg placed on mass m1 then we have

T - f = (m1 + 1.2)a2    ............(3)

The acceleration of the masses

S =0.5 a2 t2

a2 =2S / t2

= 2*1.5 / 1.3 *1.3

= 1.78 m/s2

Subtracting 2 and 3 we get

(m1 + 1.2 ) a2 = m1a1

(m1 + 1.2) * 1.78 = m1 * 5.95

m1 ( 5.95 - 1.78) = 1.2 * 1.78

m1 = 0.51 Kg

--------------------------------------

T - f = m1a1

f = T - m1a1

mu k m1g = T -m1a1

mu k = (T - m1a1) /m1g

= ( 9.63 - 0.51 * 5.95 ) / 0.51*9.8

= 1.32

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