The driver of a speeding empty truck slams on the brakes and skids to a stop through a distance d. On a second trial, the truck carries a load that doubles its mass. What will now be the truck's skidding distance?
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the same
stopping distance can be found from
vf^2=V0^2 + 2ad where a is the acceleration
the acceleration is due to friction, so we have for the force
F = m a = u mg or a = u g
all that matters is the coefficient of friction and gravity; if you double the mass, you also double the normal force and double the force of friction, so the effects of mass cancel out
hence, for the same conditions and initial velocity, acceleration is the same and stopping distance is the same
stopping distance can be found from
vf^2=V0^2 + 2ad where a is the acceleration
the acceleration is due to friction, so we have for the force
F = m a = u mg or a = u g
all that matters is the coefficient of friction and gravity; if you double the mass, you also double the normal force and double the force of friction, so the effects of mass cancel out
hence, for the same conditions and initial velocity, acceleration is the same and stopping distance is the same
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initial case:
0.5 m v^2 = F * d
d = mv^2 / 2F
F: braking force
final case:
0.5 * 2m v^2 = F * D
D = mv^2 / F = 2d
answer
truck's skidding distance = 2d
answer
0.5 m v^2 = F * d
d = mv^2 / 2F
F: braking force
final case:
0.5 * 2m v^2 = F * D
D = mv^2 / F = 2d
answer
truck's skidding distance = 2d
answer
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The distance will be the same as when unloaded.
a = µ*g,
which shows that the acceleration is not a function of mass, but only µ and the local value of g.
a = µ*g,
which shows that the acceleration is not a function of mass, but only µ and the local value of g.