Question

A block of mass 2 kg is lowered by 10 cm using a rod starting from rest. Its speed at lowest point is 2 ms-1. Work done by tension in the rod is
(Take g = 10 ms-2)

(1) Zero
(2) 2J
(3) 4J
(4) 6J
​

Answer 1

Answer:

2 J

Explanation:

m = 2 kg

d = 10 cm = 0.1 m

Initial velocity Vo = 0 m/s

final velocity V = 2 m/s

Let us assume that force applied by rod is F.

now gravitational force = mg = 2*10 = 20 N

since F is lowering the block we assume that it is acting downwards. so the net force on the block is F + 20 N.

acceleration, a = (F+ 20)/m = (F+20)/2

now we know that V^2 - (Vo)^2 = 2*a*d. putting values,

2^2 - 0 = 2*[(F+20)/2]*0.1

4 = 0.1 *( F+20 )

F = 20 N

Now, work = force*distance = 20*0.1 = 2 J

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