Question

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.

Answer 1

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.