1) A car moving along a straight stretch of road at 60 mph [26.7 meters/sec]. The driver makes slams on his breaks locking them so that the car skids to a stop. Assume a constant braking force due to friction, and a kinetic friction coefficient of 0.70 between the tires and the road: (a) calculate the braking distance from the instant that the brakes were applied, (b) how much work did the friction do in stopping the car?
vf^2=vi^2+2ad
where a= force/mass=-.7mg/m= -.7g
solve for distance d
To calculate the braking distance and the work done by friction, we need to use the following equations:
1) Braking distance (d): d = (v^2) / (2u*g)
2) Work done by friction (W): W = F * d
Where:
- v is the initial velocity of the car in m/s (26.7 m/s in this case)
- u is the kinetic friction coefficient between the tires and the road (0.70 in this case)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- F is the force of friction, which can be calculated using the equation F = u * m * g
- m is the mass of the car (we'll assume it to be 1,000 kg for this example)
(a) Calculating the braking distance:
Using the equation d = (v^2) / (2u*g), we can substitute in the given values:
d = (26.7^2) / (2 * 0.70 * 9.8)
d = 720.92 meters
Therefore, the braking distance from the instant the brakes are applied is approximately 721 meters.
(b) Calculating the work done by friction:
To calculate the force of friction (F), we can use F = u * m * g:
F = 0.70 * 1000 * 9.8
F = 6860 N
Substituting the values of F and d into the equation W = F * d, we get:
W = 6860 * 721
W = 494 thousand Joules (or Nm)
Therefore, the work done by friction in stopping the car is approximately 494,000 Joules (or Nm).