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  • Confused! kindly explain, A force acts on a 2kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds?

A force acts on a 2kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds?

 

  • Option 1)

    850 J

  • Option 2)

    875 J

  • Option 3)

    950 J

  • Option 4)

    900 J

Answers (1)

best_answer

 

Definition of work done by variable force -

W=\int \vec{F}\cdot \vec{ds}

- wherein

\vec{F} is variable force and \vec{ds} is small displacement

Work done = change in kinetic energy

V=\frac{dx}{dt}=6t

 Work\: \: done =\frac{1}{2}mv^{2}-0

                           =\frac{1}{2}(2)(6\times 5)^{2}-0

                          =900J

                      

                  

 


Option 1)

850 J

Option 2)

875 J

Option 3)

950 J

Option 4)

900 J
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