Question

An initially stationary 2.7 kg object accelerates horizontally and uniformly to a speed of 13 m/s in 4.0 s. (a) In that 4.0 s interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?

Answer 1

Explanation:

A.

Given:

V = 13 m/s

t = 4 s

Constant acceleration, a= (V-Vi)/t

= 13/4

= 3.25 m/s^2

F = mass * acceleration

= 2.7 * 3.25

= 8.775 N.

Using equations of motion,

distance,S = (13 * 4) - (1/2)(3.25)(4^2)

= 26 m

Workdone, W = force * distance

= 8.775 * 26

= 228.15 J

B.

Instantaneous power, P = Force *Velocity

= 8.775 * 13

= 114. 075 W

C.

t = 2 s,

Constant acceleration, a= (V-Vi)/t

= 13/2

= 6.5 m/s^2

Force = mass * acceleration

= 2.7 * 6.5

= 17.55 N

Instantaneous power, P = Force *Velocity

= 17.55 * 13

= 228.15 W.

= 114. 075 W.