Answer to Question #4241 in Mechanics | Relativity for neha
x=3t^2,where x is in metre and time t is in second.If mass of the body
2kg ,then find the instantaneous power delivered to body by force
on it at t=4sec.
First notice that
of a body, & v = x' = 6t m/s
& acceleration of a body, a = v' = 6
Hence the force acting on the body is
F = ma = 2 * 6 = 12
Since F is constant, then the work of this force to move the
from the point& 0& to& x& is equal to
A = F x,
instantaneous power delivered to body by force is
p = A' = F x' = F
Then at t=4sec we will have
p(4) = F * v(4) = 12 * 6 * 4 = 288
Answer: 288 W.
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