a. V^2 = Vo^2 + 2g*h.
V^2 = 0 + 19.6*5 = 98,
V = 9.90 m/s.
b. Fn = Mg*Cos A = 2.25*9.8*Cos 0 = 22.1 N. = Normal force.
A small object of mass m = 2.25 kg is released, starting at rest, from a height h above the ground on a ramp inclined at 45.0o (see figure below). The object reaches the bottom of the ramp and
enters a loop-the-loop of radius R = 1.25 m. There is no friction between the object and the track. Treat the object as a point mass.
a) What is the speed of the object at position B, the top of the loop, if h = 5.00 m?
b) What is the magnitude of the normal force acting on the object when it is at position B,
the top of the loop, having started at the height h = 5.00 m? Be sure to draw a free body
diagram for m at the top of the loop.
c) Find the minimum starting height, hmin, for which the object will just make it through the
loop without leaving the track at B. (Hint: Consider what happens to the normal force at the top of the loop when the object just makes it through the loop).
1 answer