Question

An initially stationary 2.3 kg object accelerates horizontally and uniformly to a speed of 8 m/s in 3.0 s.(a) In that 3.0 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?(c) At the end of the first half of the interval?

Answer 1

Answer:

a) 147.4 J

b) 49.13 W

c) 98.27 W

Explanation:

a) The work is placed when a force moves an object. In horizontally, it is:

W = F.d, where W is the work, F is the force and d the distance.

The force is F = m*a, where m is the mass of the object and  the acceleration, which is:

a = v/t, where v is the velocity, and t is the time, so:

a = 8/3 = 2.67 m/s²

The distance is

d = v*t

d = 8*3

d = 24 m

W = 2.3*2.67*24

W = 147.4 J

b) The instantaneous power (P) is the work in a certain time, so:

P = W/Δt

P = 147.4/3

P = 49.13 W

c) The end of the first half of the interval will be at t = 1.5 s

P = 147.4/1.5

P = 98.27 W