A force of 4N acts on a 10 kg body initially at rest. Let W_1 is work done by force during 0 t 1s. Likewise W_2 is the work done by force during 1s t 2s, where t is time in second. The ratio W_2W_1,…

Question

A force of 4N acts on a 10 kg body initially at rest. Let W_1 is work done by force during 0 t 1s. Likewise W_2 is the work done by force during 1s t 2s, where t is time in second. The ratio W_2W_1, is

Accepted Answer

Correct answer: D. (d) Acceleration because of applied force, a= Fm= 410=0.4 ms^-2 In interval 0 t 1s ,t^ '=1s s_1=ut_1^1+ 12at^1=0 1+ 12 0.4 1^2 ∴ Work done is given as =0.2m W_1=F s_1 In interval 1 t 2s ,t^ ' '=1s=4 0.2=0.8J Velocity of the body after t^ '=1s is v, then v=0+at_1=0+0.4 1=0.4 ms^-1s_2=vt_2+ 12at_2^2=0.4 1+ 12 0.4 1^3=0.6m ∴ Work done, W_2=F_2=4 0.6=2.4J ∴ W_2W_1= 2.40.8= 31=3

$A force of 4 N acts on a 10 kg body initially at rest. Let W_{1} is work done by force during 0 \leq t \leq 1 s . Likewise W_{2} is the work done by force during 1 s \leq t \leq 2 s, where t is time in second. The ratio \frac{W _{2}}{W _{1}}, is$

Solution

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