Solving a Power Problem: Work & Inst. Power

[AnkhUNC]

[SOLVED] Power Problem

Homework Statement

An initially stationary 2.9 kg object accelerates horizontally and uniformly to a speed of 14 m/s in 4.7 s. (a) In that 4.7 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?

Homework Equations

The Attempt at a Solution

Not really sure what's involved in this question. I need to find Power which is F*velocity but I don't know the force. Force = m*a. What I am given is the mass and speed is the velocity right? So I need to solve for a. Is this correct?

 

[Saladsamurai]

Force = m*a. What I am given is the mass and speed is the velocity right? So I need to solve for a. Is this correct?

Sounds good to me.

 

[AnkhUNC]

So finding a via v=v0+at I get a = 2.978723404. So for a it should be m*a = 8.638297872 but this is wrong. What did I do wrong :(?

 

[Saladsamurai]

So finding a via v=v0+at I get a = 2.978723404. So for a it should be m*a = 8.638297872 but this is wrong. What did I do wrong :(?

a Asks you for WORK. What is the definition of work done by a constant force?

 

[AnkhUNC]

Nono I meant a = acceleration. I get 8.638 N for the force.

 

[AnkhUNC]

I get the following:
a = 2.979; F = ma = 2.9*2.979 = 8.638 N

V1 = a*t/2 = 2.979*(4.7/2) = 7.000 m/s
V2 = 14.000 (given)

P1 = F*V1 = 8.638*7 = 60.466 W = (c)
P2 = 8.638*14 = 120.932 W = (b)

But it was (a) to be in Joules. However when I multiply my F in Newtons by meters (14*4.7) the answer isn't correct :( but (b) and (c) are.

 

[AnkhUNC]

W = KE = ½mV² = ½*2.9*14² = 284.2 J