Question

A man of 60 kg weight is standing at rest on a platform. He jumps up vertically a distance of 1 m and the platform at the same instant moves horizontally forward with the result that the man lands 1 meter behind the point on the platfrom from where the took the jump the total work done by the man at the instant he lands is
(a) 300 J
(b) 150 J
(c) 600 J
(d) zero
Answer
Answer: (c) 600 J

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Answer 1

The work done is 600 J.

Option (c) is correct.

Explanation:

We are given that:

• Mass of man = 60 Kg
• Distance covered by him as he jumps up = 1 m
• To Find: The work done by him when he lands on platform = ?

Solution:

The formula of work done is

Work = force x displacement

W = F x d

Now Force "F" = mg

W = mg x d

W = 60 x 10 x 1

W = 600 J

Thus the work done is 600 J

Answer 2

Answer:

We are given that:

Mass of man = 60 Kg

Distance covered by him as he jumps up = 1 m

To Find: The work done by him when he lands on platform = ?

Solution:

The formula of work done is

Work = force x displacement

W = F x d

Now Force "F" = mg

W = mg x d

W = 60 x 10 x 1

W = 600 J

Thus the work done is 600 J

Explanation: