Question

Q. Two unequal masses of 1 kg and 2 kg are attached at the two ends of a light inextensible string passing over a smooth pulley as shown in figure. If the system is released from rest, find the work done by string on both the blocks in 1 s. (Take g = 10 m/s²)​

Answer 1

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explanation

Answer 2

$\huge\fcolorbox{black}{magenta}{Solution:-}$

✏️ Answer:-

★ Work done by tension on both 1 kg and 2 kg blocks in a given time period of 1 s are 200/9 J and -200/9 J respectively.

✏️ Rationale:-

★ As per the portrayed scenario,

The net pulling force acting on the system is

Fⁿᵉᵗ = 2g - 1g

⇒Fⁿᵉᵗ = 20 - 10

⇒Fⁿᵉᵗ = 10 N

Now,

The total mass being pulled, m = ( 1 + 2 ) = 3 kg.

Accordingly,

Acceleration of the system, will thus be,

a = Fⁿᵉᵗ/m

⇒a = 10/3 m/s²

Displacement of both the blocks in midst of 1s is -

S = 1/2 at²

⇒S = 1/2 [10/3] × [1]²

⇒S = 5/3 m.

Now, as of the scene,

From the free body diagram of 2 kg block, [Kindly have a glimpse at the attachment provided herewith!]

Using ΣF = ma, we acquire,

20 - T = 2a = 2 [10/3]

Or,

T = 20 - 20/3

⇒T = 40/3 N

∴ Work done by the string (in accordance with the tension) on 1 kg block in 1 s will be,

W1 = (T) (S) ( cos 180° )

Hence,

On plugging in the values of the units from the question, we get,

W1 = [40/3] × [5/3] × [1]

⇒W1 = 200/9 J

In a similar manner,

Work done by the string on 2 kg block in a given period of 1s will be,

W2 = (T) (S) (cos 180°)

⇒W2 = [40/3] × [5/3] × [-1]

⇒W2 = - 200/9 J

∴ Work done by tension on both 1 kg and 2 kg blocks in a given time period of 1 s are 200/9 J and -200/9 J respectively!

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I hope this helps! :)