KVPY SA Exam 2011 Question Paper

Question : 67 of 80

Marks: +1, -0

A body of 0.5 kg moves along the positive x-axis under the influence of a varying force F (in Newtons) as shown below.

If the speed of the object at x = 4 m is 3.16 m s

^{-1}

then its speed at x = 8 m is

Solution:

According to work-energy principle

W_C + W_{nc} + W_{ext }= \∆KE

∫↖{8}↙{4}\Fdx=1/2mv_f^2-1/2mv_i^2

⇒1/2×3×8-1/2×1.5×4

=1/2×1/2[v_f^2-(3.16)^2]

v_f

= 6.8 m/s

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