Guys I'm confused with this, different answers all over the place. The 'correct' answer does not make much sense either.
>A speeding truck locks it brakes and it skids to a stop. If the truck's total mass were doubled,what would happen to its skidding distance? Why?
If you work it out mathematically it would be 1/2 the original stopping distance.
But the 'official answer' is they are the same? Are they just accounting for the brakes? Don't brakes rely on friction, The answer should be 1/2 of the before mentioned skidding distance. Help /sci?
Using basic mechanics you can get mass to divide out on both sides of the equation. This means that the mass doesn't matter, so the time will be the same.
Ffriction = Fnormal * mu
Fnormal = weight
substitute known quantities
>If you work it out mathematically it would be 1/2 the original stopping distance.
What? No it wouldn't. You fucked something up. If it's twice the mass it has twice the kinetic energy, since the frictional force of the brakes is assumed to be the same in both trucks, and force*distance is work, the work needed to cancel the KE will be twice the distance times the force (which remains the same)
Should be the same, based on the high-school physics analysis. Frictional force doubles because the normal force doubles, but inertia doubles as well, so by F = ma the deceleration should be the same.
You should assume that the vehicle has the same speed in both cases, and that in both cases the brakes immediately stop the wheels from turning. Since momentum and friction force are both proportional to mass, changing the mass won't change the skidding distance.
In reality, the tires are fairly plastic and thus the coefficient of friction will also be dependent on mass, as well as the strength of the deceleration which just compounds the stopping force.
So a heavy truck does indeed stop faster.
Except in the case where the tires overheat or even fail catastrophically, at which point the coefficient of friction drops significantly and then the heavier truck takes much longer to come to a halt.
Kinetic energy of the "double truck" : m*v^2
Where m is the mass of the simple truck.
Let D be the stopping distance of the double truck, and d the stopping distance of the simple truck.
Ft is the tangential force of the wheels on the road for the simple truck, this force being proportional to the mass (Ft = Fn*f = g*m*f) the tangential force for the double truck is 2*Ft.
d*Ft = 0.5*m*v^2 (1)
D*2*Ft = m*v^2 (2)
D*Ft = 0.5*m*v^2
D*Ft = d*Ft (substitute with 1)
D = d