The "Inclined Plane" Truth - solutions

46. This problem is the same as our do-now, though it's a little hidden. Stopping distance refers to the distance the block travels as it slows to a stop.
b) 47.37 meters (don't forget to convert 95 km/h into m/s!)
c) the acceleration becomes (1/6) the acceleration due to gravity. If a = 1.63 m/s^2, then the new distance is 284.2 meters.

49.
a) If there is no friction, then a = g sin(theta). If we assume a distance d along the slide (it can be anything!) then we can derive that the speed the child has at the bottom of the slide is V_no-friction=(2gd sin(theta))^(1/2).

If there is friction, the problem says the speed at the bottom is HALF this quantity.
With friction, solve for the new acceleration, and solve for the final velocity of the child at the bottom of the slide. The displacement is the SAME as with no friction.

You should get that a = g sin(theta) - mu*g*cos(theta) and that the final velocity is:
(2gd*sin(theta)-2*mu*gD*cos(theta)) ^ (1/2). If you take this expression and set it equal to 0.5*V_no-friction, you should be able to solve for mu.

The answer (phew!) is that mu = 0.75*tan(28) == .399


51.
a) 1.792 seconds
b) The acceleration does NOT depend on the mass, so the time required would be the same.

52.
a) a = g sin(theta) = 3.67 m/s^2
b) v = (2*a*d)^1/2 = 8.17 m/s

53.
a) 1.226 meters
b) 1.634 seconds (This is just like finding the total time a projectile is in the air - find the time required for the block to stop, and then double it for the entire trip.)

54.