A force of 40 N acts on body of mass 5 Kg which is initially at rest. What is the amount of work done in the first 10 s?(a) 16000 J(b) -16000 J(c) 400 J(d) -400 JThis question was addressed to me in an interview for internship.The above asked question is from Work in chapter Work, Energy and Power of Physics – Class 11

A force of 40 N acts on body of mass 5 Kg which is initially at rest.

1.	A force of 40 N acts on body of mass 5 Kg which is initially at rest. What is the amount of work done in the first 10 s?(a) 16000 J(b) -16000 J(c) 400 J(d) -400 JThis question was addressed to me in an interview for internship.The above asked question is from Work in chapter Work, Energy and Power of Physics – Class 11
Answer» The CORRECT answer is (a) 16000 J To explain: The work done on a body is obtained by CALCULATING the DOT product of the force and displacement vector. From Newton’s second law, we get the acceleration EQUAL to 8 m/s^2. From the second equation of motion the displacement COVERED can be obtained which is equal to 400 m. On taking the dot product of the force and displacement, we get work done equaling to 16000 J.

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