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Standard XII

Physics

Speed of Efflux

Q. Under the ...

Question

Q. Under the action of a force , a 2 kg body moves such that its position x as a function of time t is given by , x=t^2/3 , where x is in metre and t in seconds. The work done by the force in first two seconds is ?

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Solution

work=forcedispacement
w=fs
f=ma m=2
x=t^2/3
displacemet at t=2 is x=2^(2/3)................(1)
v=dx/dt=2/3t^((2/3)-1)
=(2/3)t^(-1/3)
a=dv/dt=(2/3)(-1/3)t^((-1/3)-1)
a=(-2/9)t^(-4/3)
acceleration at t=2 is
a=(-2/9)2^(-4/3)......(2)
therefore w=mas=2(-2/9)2^(-4/3)*2^(2/3)
on solving
w=-0.28 joule

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Q. Under the action of a force , a 2 kg body moves such that its position x as a function of time t is given by , x=t^2/3 , where x is in metre and t in seconds. The work done by the force in first two seconds is ?

work=force*dispacement w=fs f=ma m=2 x=t^2/3 displacemet at t=2 is x=2^(2/3)................(1) v=dx/dt=2/3t^((2/3)-1) =(2/3)t^(-1/3) a=dv/dt=(2/3)(-1/3)t^((-1/3)-1) a=(-2/9)t^(-4/3) acceleration at t=2 is a=(-2/9)2^(-4/3)......(2) therefore w=mas=2*(-2/9)2^(-4/3)*2^(2/3) on solving w=-0.28 joule

work=forcedispacement
w=fs
f=ma m=2
x=t^2/3
displacemet at t=2 is x=2^(2/3)................(1)
v=dx/dt=2/3t^((2/3)-1)
=(2/3)t^(-1/3)
a=dv/dt=(2/3)(-1/3)t^((-1/3)-1)
a=(-2/9)t^(-4/3)
acceleration at t=2 is
a=(-2/9)2^(-4/3)......(2)
therefore w=mas=2(-2/9)2^(-4/3)*2^(2/3)
on solving
w=-0.28 joule