Question #4241

The position(x) of a body moving along x-axix at time (t) is given by x=3t2,where x is in metre and time t is in second.If mass of the body is 2kg ,then find the instantaneous power delivered to body by force acting on it at t=4sec.

Expert's answer

The position(x) of a body moving along x-axix at time (t) is given

by

x=3t^2,where x is in metre and time t is in second.If mass of the body

is

2kg ,then find the instantaneous power delivered to body by force

acting

on it at t=4sec.

Solution.

First notice that

& velocity

of a body, & v = x' = 6t m/s

& acceleration of a body, a = v' = 6

m/s^2

Hence the force acting on the body is

F = ma = 2 * 6 = 12

N

Since F is constant, then the work of this force to move the

body

from the point& 0& to& x& is equal to

A = F x,

whence the

instantaneous power delivered to body by force is

p = A' = F x' = F

v.

Then at t=4sec we will have

p(4) = F * v(4) = 12 * 6 * 4 = 288

W.

Answer: 288 W.

by

x=3t^2,where x is in metre and time t is in second.If mass of the body

is

2kg ,then find the instantaneous power delivered to body by force

acting

on it at t=4sec.

Solution.

First notice that

& velocity

of a body, & v = x' = 6t m/s

& acceleration of a body, a = v' = 6

m/s^2

Hence the force acting on the body is

F = ma = 2 * 6 = 12

N

Since F is constant, then the work of this force to move the

body

from the point& 0& to& x& is equal to

A = F x,

whence the

instantaneous power delivered to body by force is

p = A' = F x' = F

v.

Then at t=4sec we will have

p(4) = F * v(4) = 12 * 6 * 4 = 288

W.

Answer: 288 W.

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satyam24.09.11, 08:48ANSWER 144

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