Question

A force acting on a body of mass 2kg moves it, such that its position y as a function of time is given by the relation y = t³/3; where y is in metre and t is in sec. Calculate the work done by the force in first two second.​

Answer 1

Answer:

ANSWER

v=

dt

dx

=t

2

W=

2

1

mv

2

=

2

1

mt

4

=

2

1

×2×(2)

4

=16J