A 4.0 kg block, lying on a smooth table, is connected to a 5.0 kg block by means

Question

A 4.0 kg block, lying on a smooth table, is connected to a 5.0 kg block by means of a weightless cord passing over, without slipping, a cylindrical pulley of 2.0 kg mass and 10.0 cm radius placed at the edge of the table. The system is released from rest at t = 0. In 4.0 s the 5.0 kg block will drop by The tension in the part of the cord connected horizontally to the 4.0 kg block is A uniform. 1.0 m radius, 50.0 kg circular wheel (I = M R^2) spinning at 120 rev/min is subjected to a variable frictional force F(t) = 20.0[1 - (t / lambda)^2] applied tangentially at its rim. If the wheel has to stop rotating exactly at the moment the force becomes zero, the value of lambda must be The number of revolutions done by the wheel before coming to rest is

Explanation / Answer

4
(a)
Let the tenison in the string connected to 5.0 Kg block = T1
Let the tenison in the string connected to 4.0 Kg block = T2
Let the Acceleration of the system = a

m1 * a = m1*g - T1
5.0 * a = 5.0 * g - T1
T1 = 5.0 *g - 5.0*a --------1

m2 * a = T2 - m2*g
4.0 * a = T2 - 4.0 * g
T2 = 4.0 * a + 4.0 * g ----------2

Now,
(T1 - T2)*r = I*
Where = a/r
(T1 - T2)*r = I* a/r
(T1 - T2) = I*a/r^2

I = 1/2 * m*r^2
(T1 - T2) = 1/2 * m*r^2 *a/r^2
(T1 - T2) = 1/2 * m*a
T1 = 1/2 *ma + T2
T1 = a +T2

5.0 *g - 5.0*a = a + 4.0 * a + 4.0 * g
g = 10.0 * a
a = 0.98 m/s^2

S = u*t + 1/2 * at^2
S = 0 + 1/2 * 0.98 * 4^2 m
S = 7.84 m

(b)

T2 = 4.0 * a + 4.0 * g
T2 = 4.0 * 0.98 + 4.0 * 9.8
T2 = 43.12 N

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