If a 12 kg12kg object is constantly accelerated from 0m/s0ms to 40 m/s40ms over 8 s, how much power musy be applied at t=1 t=1?

1 Answer
Mar 30, 2017

300"W"300W (Watts)

Explanation:

The power applied to an object is defined as P=W/tP=Wt, and expanding the definition of work, P=(Fd)/tP=Fdt (assuming force and displacement are in the same direction). d/t=vdt=v, so P=FvP=Fv. F=maF=ma, so P=mavP=mav.

The acceleration of an object is (vec(v_f)-vec(v_i))/(Deltat), for this object it is (40"m/s"-0"m/s")/(8"s")=5"m/s"^2.

The velocity of an object at time t is at, so in this case the velocity at 1"s" is 5"m/s"^2*1"s"=5"m/s"

Finally, solving for power using P=mav, we get P=12"kg"*5"m/s"^2*5"m/s"=300"W"