a = F / m = -9000 / 1500 = -6m/s^2.
d = (Vf^2 - Vo^2) / 2a,
d = (0 - (20)^2) / -12 = 33.33m.
A force of -9000 N is used to stop a 1500kg car traveling at 20 m/s. What braking distance is needed to bring the car to a halt?
2 answers
Kinetic energy of car before breaking:
Wk = mass * velocity^2/2
Wk = 1500 * 20^2/2
The work done by the car to transfer kinetic energy to frictional heat:
W(friction) = Force * Distance
-9000n * Distance
As no energy can be lost only transformed, we can write: 1500 * 20^2 /2 = -9000n * distance
solving for distance we get:
Distance = 1500 * 20^2/2 * -9000
Distance = -33,333
As distance can't be negative and is a scalar quantity we can write 33.333m
Wk = mass * velocity^2/2
Wk = 1500 * 20^2/2
The work done by the car to transfer kinetic energy to frictional heat:
W(friction) = Force * Distance
-9000n * Distance
As no energy can be lost only transformed, we can write: 1500 * 20^2 /2 = -9000n * distance
solving for distance we get:
Distance = 1500 * 20^2/2 * -9000
Distance = -33,333
As distance can't be negative and is a scalar quantity we can write 33.333m
0 answers