Physics Question: EK 1001 #226

[grey_wind]

A car moving at 20 m/s brakes and slides to a stop. If the coefficient of kinetic friction between the pavement and the tires of the car is 0.1, how far does the car slide?

A. 50m
B. 100m
C. 200m
D. 400m

This is the explanation in the back of the book, but I'm still lost... please spell everything out for me, long day haha, thanks!

C is correct. The net force on the car is the kinetic frictional force umgcos(theta). From F=ma we can find the acceleration and plug this into v^2 = vo^2 + 2ax. You could also solve this using the energy formula: friction*d = 1/2mv^2

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[csx]

look up work energy thereom. That should explain the usage of the second method.

 

[blu3]

Work friction = uk(Fn)(d)
KE = 1/2mv^2

W = delta KE
ukmgd = 1/2mv^2

m will cancel out.
solve for d and you get 200

 

[TactFR44]

A car moving at 20 m/s brakes and slides to a stop. If the coefficient of kinetic friction between the pavement and the tires of the car is 0.1, how far does the car slide?

A. 50m
B. 100m
C. 200m
D. 400m

This is the explanation in the back of the book, but I'm still lost... please spell everything out for me, long day haha, thanks!

C is correct. The net force on the car is the kinetic frictional force umgcos(theta). From F=ma we can find the acceleration and plug this into v^2 = vo^2 + 2ax. You could also solve this using the energy formula: friction*d = 1/2mv^2

Ok. So if you draw out your force diagram, you place Ff = MA. The only force in the X (the force that matters here, is the kinectic friction that is stopping the car). You know it's kinectic because the the car is SLIDING to a stop. Two surfaces sliding = kinectic friction in the works yo.

Once you have this, then lets solve for the acceleration.

Ff=mA
Fn*uk=ma
Mg*uk=ma
G*uk=a
a=1m\s^2

Now, kinmetics yo. You have 3 variables. velocity intial = 20, a=1 vf=0.
So, which equation since we want delta x?

the one without t!
Vf^2=vi^2+2adeltax.
breaking it down and plug and chug it. Vi^2/2a=deltha x
400/2= 200m or choice c

 

[TactFR44]

Ok. So if you draw out your force diagram, you place Ff = MA. The only force in the X (the force that matters here, is the kinectic friction that is stopping the car). You know it's kinectic because the the car is SLIDING to a stop. Two surfaces sliding = kinectic friction in the works yo.

Once you have this, then lets solve for the acceleration.

Ff=mA
Fn*uk=ma
Mg*uk=ma
G*uk=a
a=1m\s^2

Now, kinmetics yo. You have 3 variables. velocity intial = 20, a=1 vf=0.
So, which equation since we want delta x?

the one without t!
Vf^2=vi^2+2adeltax.
breaking it down and plug and chug it. Vi^2/2a=deltha x
400/2= 200m or choice c

Or do exactly what the poster above me did! Easier and faster. I explained utilizing the method EK used.

 

[grey_wind]

Or do exactly what the poster above me did! Easier and faster. I explained utilizing the method EK used.

were both very helpful! thanks.

Just a little burnt out today haha, the EK explanation mentioned Cos theta and that tripped me up

 

[SN2ed]

Thread closed. These types of questions belong in the MCAT Study Q&A forum only.