Register to reply

Work and Power

by g.uricchio
Tags: power, work
Share this thread:
g.uricchio
#1
Feb4-09, 07:53 PM
P: 5
1. The problem statement, all variables and given/known data

A car of mass 840.0 kg accelerates away from an intersection on a horizontal road. When the car speed is 37.4 km/hr (10.4 m/s), the net power which the engine supplies is 4300.0 W (in addition to the extra power required to make up for air resistance and friction). Calculate the acceleration of the car at that time.

2. Relevant equations

Work=Force*Distance

Power=Work/Time

Force=Mass*Acceleration

3. The attempt at a solution

I have tried numerous ways of solving this problem and remain unsuccessful. My closest attempt was using the equation Power=Kinetic Energy/Time, I used the velocity as Kinetic Energy to find the time and plugged that into the Kinematic equation Vf=Vi+a*d but that did not work either. Please help me!
Phys.Org News Partner Science news on Phys.org
Hoverbike drone project for air transport takes off
Earlier Stone Age artifacts found in Northern Cape of South Africa
Study reveals new characteristics of complex oxide surfaces
Delphi51
#2
Feb4-09, 09:49 PM
HW Helper
P: 3,394
Do you know some calculus?
It looks like a calculus problem to me.
You can't use Power = KE/time: consider the situation where the car is moving at zero acceleration and a large speed. The power required ("extra" not counted as stated in problem) is zero but the KE is large.

Instead, you should use Power = dE/dt, the derivative of kinetic energy with respect to time. The differentiation is not difficult, but does involve the chain rule.
g.uricchio
#3
Feb5-09, 09:04 AM
P: 5
yes I do know calculus, I am in Calc 2 but I still don't follow.

timon
#4
Feb5-09, 09:34 AM
P: 56
Work and Power

M= 840.0 kg
P= 4300.0 W
v= 10.4 m/s

s = v*dt [you are working with power (which is watt per second). what should dt be?]

gives s

p = w/dt

gives w

w = f*s

gives f

f = m*a
g.uricchio
#5
Feb5-09, 11:45 AM
P: 5
how am I supposed to find dt when im not given a time
LowlyPion
#6
Feb5-09, 11:57 AM
HW Helper
P: 5,343
As I recall power is equal to force times velocity.

Units of Watts = N-m/s

P = F*V

So the instantaneous power is given as 4300 w and the velocity is 10.4 m/s.

That makes F = 4300/10.4 doesn't it?

And you have the mass, so ...
g.uricchio
#7
Feb5-09, 11:59 AM
P: 5
i thought that power was equal to F*V but if you do the math it delivers units of Kg*m/s and not Kg*m/s2. maybe im using the wrong velocity?
Hannisch
#8
Feb5-09, 12:33 PM
P: 116
Watt = J/s = kgm2/s3 = Nm/s

So it works out just fine. :)
Delphi51
#9
Feb5-09, 07:51 PM
HW Helper
P: 3,394
Oh, clever to remember P = Fv !

Just for interest, I was thinking of P = dE/dt = d/dt(.5mv^2) = .5m*2v*dv/dt = mva
which is the same as P = Fv.


Register to reply

Related Discussions
Work and Power Introductory Physics Homework 4
Work, Energy and Power (Work Problem) Introductory Physics Homework 9
Work/Power Introductory Physics Homework 4
Work.Power Introductory Physics Homework 2
Work and power Introductory Physics Homework 3