Two objects with masses of 4.00 kg and 5.00 kg are connected by a light string t

Question

Two objects with masses of 4.00 kg and 5.00 kg are connected by a light string thatpasses over a frictionless pulley.
(a)Determine the tension in the string.
(b)Determine the acceleration of each object.
(c)Determine the distance each object will move in the first second ofmotion if both objects start from rest.
(a)Determine the tension in the string.
(b)Determine the acceleration of each object.
(c)Determine the distance each object will move in the first second ofmotion if both objects start from rest.

Explanation / Answer

First, we have to notice that the lighter block acceleratesupwards, so the tension is in the direction of theacceleration and the force due to gravity is against it, so the netforce is equal to T - mg. In the same way, the heavier blockaccelerates downwards, so the net force is equal to Mg - T. A) On the lighter block: F = T - mg = ma a = (T - mg)/m = T/m - g On the heavier block: F = Mg - T = Ma a = (Mg - T)/M = g - T/M Combining these gives that T/m - g = g - T/M TM - mMg = mMg - Tm T(M + m) = 2mMg T = 2mMg/(M + m) = 2(4.00 kg)(5.00 kg)(9.8m/s2)(5.00 kg + 4.00 kg) = 43.6N B) To find the acceleration, we use one of the expressions wefound earlier for the acceleration a = g - T/M = 9.8 m/s2 - (43.6N)/(5.00 kg) =1.09 m/s2 C) To find the distance they will move in one second, use theequation x = vot + at2/2 = (0)(1s) + (1.09m/s2)(1s)2/2 = 0.544m C) To find the distance they will move in one second, use theequation x = vot + at2/2 = (0)(1s) + (1.09m/s2)(1s)2/2 = 0.544m

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