How to Calculate Work and Power in a Physics Problem Involving Velocity and Time

[Cheapo2004]

I have a quick question, I need to find Work for this problem, but I don't have a distance!? The problem provides vi and vf, time, and mass, but no distance? How do i find work?

1. A 1.50 X 10^3 kg car accelerates from rest to 10m/s in 3 seconds.
A. What is the work done in this time? W=?
B. What is the power delivered by the engine? P=?

W= Force * Distance (cos theta) <-- Right?

W=? P=? m=1500 kg vi=0m/s vf=10m/s T=3 seconds

-------------------------------------------------------------
******************Edit**************************
-------------------------------------------------------------

I solved my own problem!
x = .5 (vi+vf)time

Anyone want to check my work for me?

W=? P=? m=1500 kg vi=0m/s vf=10m/s T=3 seconds x=?

x= .5 (0+10) 3
x= 15m

Force = mass * acceleration
F = 1500kg * 10...?
F = 15000 J?

W = 15000 * 15
W = 225000 J...?

Thats a lot of work!?

-------------------------------------------------------------

After somebody could nicely check my work i have one more problem...

A skateboard reaches a speed of 35m/s from an initial speed of
25m/s after 21KJ of work

I THINK what the problem is asking for here is: What is the skateboarder's kinetic energy disregarding the mass of the skateboard?

 

[mezarashi]

Unfortunately, I don't think your approach was correct. Although, you would have arrived at the same answer (given you had used the correct value for acceleration under the assumption that it is constant), I think it is better you use concepts from Work & Energy rather than mixing in Kinematics unnecessarily.

Firstly, the equation: [d = 0.5(Vf+Vi) x t] only works when the acceleration is constant. There is no reason to assume this from what is given.

Going back to the first principles of work and energy: the work done on an object over a period of time or distance is equal to the change in kinetic energy of that object.

From the question, can you calculate the kinetic energy (0.5mv^2) of the car before the acceleration? And after? What is the difference? By conservation of energy, how much energy should have been delivered?

Power is Energy/time. If that given energy above was delivered in 3 seconds, what is the power?

 

[Naumaan]

thats simple
according to ur data,
vi=0 vf=10m/s and t=3secs
u can find acceleration simply by using acceleration formula
a=(vf-vi)/t => a= 10/3 => a= 3.33m/(sec^2)
now putting it in F=m*a where m=mass
F=m*a
now u can find force
about distance u can apply
d= (vi*t) + (a*(t^2))/2
as Force and distance both are in hand
W= F*d and W can be calculated easily
after that P=W/t
so power is also calculated
hope u get the answer correct...

 

[Cheapo2004]

Ok, let's re-do this:

vi=0 vf=10m/s t=3secs m=1500 kg

a=(vf-vi)/t
a=(10)/3
a=3.3m/s^2

d= (vi*t) + (a*(t^2))/2
d= (0*3) + (3.3(9))/2
d= 14.85m

F = m * a
F = 1500 * 3.3
F = 4950

W = F * d
W = 4950 * 14.85
W = 73507.5 <--- ?

 

[Cheapo2004]

I'd love it for someone to check my latest work :!)

 

[Naumaan]

Well
ur work is gr8