You visited us 12 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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 ?

Open in App

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

Suggest Corrections

Similar questions

Q. Under the action of a force, a

2 k g

body moves such that its position

x

as a function of time is given by

x = \frac{t^{3}}{3},

where

x

is in metre and

t

in seconds. The work done by the force in the first two seconds is :

Q. Under the action of a force, a

2 k g

body moves such that its position

x

as a function of time is given by

x = \frac{t^{3}}{3}

where

x

is in metre and

t

in second. The work done by the force in the first

2 s

is :

Q. Under the action of a force, a

2 kg

body moves such that its position

x -

axis as a function of time

t

is given by

x = \frac{t^{2}}{3}

, where

x

is in metre and

t

in second. The work done by the force in first two seconds is

Under the action of force, a 2 kg body moves such that x is a function of time given by $x = t^{3} / 3$ , x in metre, t in seconds, find the work done in first two seconds.

Q. Under the action of a force, a

2 k g

body moves such that its position

x

(in meters) as a function of time

t

(in seconds) is given by

x = t^{2} / 2

. The work done by the force in the first

5

seconds is:

Join BYJU'S Learning Program

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