Physics homework, please help.?

Question

a 1200 kg car starts from rest and accelerates to 72 km/h in 20s. Friction exerts an average force of 450 N on the car during this time.
So: t=20s m=1200kg F=450N v=20 m/s (i converted it) a=9.8 m/s^2
A) What is the net work done on the car? B) How far does the car move during its acceleration?
Okay so for part A I used the equation d=d(subzero)+1/2(v:subzero+ v)t. But I overheard my teacher say that you have to square the 20. Then use W=Fd. So then I figured I could use d=1/2at^2. (the t is 20) Then use W=Fd and figure out the work.
Okay so for B) the answer is 200m. If you use d=1/2at^2 you don't get a distnace of 200. But, if you use d=d(subzero)+1/2(v:subzero+ v)t then you get a distance of 200m. But that won't make me square the 20 for part A. So...... What do I do? I have two distance formulas that don't give me the same distance.

Please show me work and what not. Thanks so much.

Answer 1

Under constant acceleration and unidirectional motion
v(t)=v0+a*t
72 km/hr = 72/3.6 m/s
v(20)=20=0+a*20
a=1

This acceleration is the net acceleration of the car which increased the kinetic energy of the car and worked against friction

The net work can be calculated a few different ways.
One, look at the increase in kinetic energy
.5*m*v^2
in this case
.5*1200*20^2
= 240 kJ

You can check this since F=m*a
the net force acting on the car is
1200*1 N
this was over a distance of 200m (incorporating the answer from part b)
again = 240 kJ

The work done by the engine of the car was (assuming purely horizontal motion)
240 +450*.2 kJ
=330 kJ
b) Again, under constant acceleration

s(t)=s0+v0*t+.5*a*t^2

The position squares the time. When you get exposed to calculus you will learn the integral of t*dt=.5*t^2

s0=0
v0=0
s(20)=.5*1*20^2
s(20)= 200 m

It looks like you assumed that the acceleration is 9.8 m/s^2, which would imply that the car was only acted on by a net force equal to g in the direction of motion.

I hope this helped you.

j

Answer 2

OHHHHHHHHH I see your problem

for b) You are using the acceleration to be 9.8.......... which is not the acceleration of the car. You have to work out the acceleration of the car using
v = v(o) + at
20 = 0 + 20a
a = 1ms(-2)

Now plug that into you distance equation for the travelling car
d = 1/2at*2
= 1/2 x 1 x (20)*2
= 1/2 x 400
= 200 m.

Tada.

FOR THE FIRST PART, you have to take into consideration the constant retardation force of 450N. That would be a negative force on your forward motion.

ALright, I got it.
Work = Energy ( so lets use Kinetic energy)
Kinetic energy = energy gained by moving car - energy lost due to friction
KE = 1/2 m( car) v*2 - 1/2 m( friction)v*2
since 1/2 v*2 is constant we take it out

ke = 1/2v*2( 1200 - 450).......... ( F = 450, but a=1 so m= 450)
= 1/2 x 400(750)
= 150,000J

Now lets work out the distance using W = Fd
Remember F=ma and in forward motion for this question, a= 1)
150,000 = (1200 -450) d
150,000= 750d
200 = d................. which is the same answer you get in PART b.

Answer 3

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Answer 4

F (net)=mass 'm'*acceleration'=m(change in velocity/time't')
=1200*(20 - zero)/20=1200N
Assuming acceleration to be constant,acceleration=(20-0)/20=1m/s^2
using d=(1/2)at^2 we get d=(1/2)*!*20*20=200 m

By second method,
Distance 'd'=(average velocity)*time={ (20+zero)/2}20=200m
Net work W=F(net)*d=240000 joule