A car traveling along a level road at speed v = 10 m/s slams on the brakes and skids to a stop. If the force of friction on the car is half the car's weight, how far does the car slide? (Hint: Use the work-energy theorem and solve for d.)
2007-03-02 06:55:42 · 3 answers · asked by Dirck G 1 in Science & Mathematics ➔ Physics
Yes, that's a great hint, using the work energy theorem :
Final energy minus initial energy is the positive value of the work made by the friction.
FInal energy will be cero, why ?, because the car stops, so the final speed will be cero.
the initial speed is 10 m/s, then :
Initial energy : mass of the car * 100 / 2 = 50*m
m = mass of the car
Weight of the car = m*g
Force of the friction = m*g/2
The work made by the force of friction : mg/2*Distance
then the final energy minus the initial energy = 50m
50m = m*g/2*D
where "d" is the distance that the car slides.
so : 50 = g/2*D
Considering g = 10 m/s^2
D = 10 meters
2007-03-02 07:00:14 · answer #1 · answered by anakin_louix 6 · 0⤊ 0⤋
You left out the coefficient of friction ot the road surface. COF ranges from 0.0 - 1.0. To figure deceleration you need to know the CoF between the tires and road surface. That is unless you presume its 1.0
2007-03-02 07:13:41 · answer #2 · answered by rico3151 6 · 0⤊ 0⤋
it is solved somewhat making use of potential equations. The kinetic potential interior the shifting vehicle is dissipated as artwork performed by potential of kinetic friction. a million/2*m*v**2 = u*m*g*d a million/2*v**2 = u*g*d u = (v**2)/(2*g*d) = (14**2)/(2*9.8*23.2) u = 0.40 3
2016-10-17 02:59:37 · answer #3 · answered by ? 4 · 0⤊ 0⤋