What force is required to stop a 1500-kg car in a distance of 0.20 m if it is initially moving at 2.2 m/s?
4 answers
Friction...
work done by force = F d = 0.20 d Joules
kinetic energy = (1/2) m v^2
work done to stop = kinetic energy
0.20 d = (1/2)(1500)(2.2)^2
kinetic energy = (1/2) m v^2
work done to stop = kinetic energy
0.20 d = (1/2)(1500)(2.2)^2
V^2 = Vo^2 + 2a*d.
0 = (2.2)^2 + 0.4a,
a = -12.1 m/s^2.
F = M*a = 1500*(-12.1) = -18,150 Joules.
The negative sign means the force opposes the motion.
0 = (2.2)^2 + 0.4a,
a = -12.1 m/s^2.
F = M*a = 1500*(-12.1) = -18,150 Joules.
The negative sign means the force opposes the motion.
banda
2 answers