Question

A weight of 290N and another of 200N are suspended by a rope on either side of a frictionless pulley. The acceleration of the system is what?
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Answer 1

Well, the tension is same all over the rope. Let T.

Let the acceleration of the system, a, act downwards for the 290N weight and upwards for the 200N weight.

Let g = 9.8 m s^(-2).

So the mass of 290N weight is 290 / 9.8 kg and that of 200N weight is 200 / 9.8 kg.

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On considering FBD of the 290N weight,

290 - T = 290a / 9.8 → (1)

On considering FBD of the 200N weight,

T - 200 = 200a / 9.8 → (2)

Adding (1) and (2),

490a / 9.8 = 90

50a = 90

a = 1.8 m s^(-2)

Answer 2

Answer:1.8 m/s

Explanation:

290 - T = 290a / 9.8 → (1)

On considering FBD of the 200N weight,

T - 200 = 200a / 9.8 → (2)

Adding (1) and (2),

490a / 9.8 = 90

50a = 90

a = 1.8 m s^(-2)