Question
A particle of mass m moves with a variable velocity v, which changes with distance covered x along a straight line as v=k√x,where k is a positive constant . The work done by all the forces acting on the particle, during the first t second is
Answer: Option C
:
C
GIven v=k√x or dxdt=k√x or x−12dx=kdt
Integrating both sides, we get
x1212=kt+C; Assuming x(0) = 0
Therefore, C = 0
2√x=kt⇒x=k2t24 or v=k2t2
Therefore, work done,
△W = Increase in KE
= 12mv2−12m(0)2=12m[k2t2]2=mk4t28
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:
C
GIven v=k√x or dxdt=k√x or x−12dx=kdt
Integrating both sides, we get
x1212=kt+C; Assuming x(0) = 0
Therefore, C = 0
2√x=kt⇒x=k2t24 or v=k2t2
Therefore, work done,
△W = Increase in KE
= 12mv2−12m(0)2=12m[k2t2]2=mk4t28
Was this answer helpful ?
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