Question

1. A motorcycle with a mass of 92 kg is inside of a vertical loop with a radius 15 meters. At the bottom the only forces acting motorcycle in the normal force and weight force. At the bottom, the amount of the normal force is 2.5 times the motorcycle's weight. As it moves to the side of the circle, some other forces act on it. But when it reaches the side, only the normal force and the weight are acting on it and the amount of the normal force on the side is 3 times the motorcycle's weight. How much net work, in Joules, is done by all forces acting on the motorcycle as it moves from bottom to the side of the circle?

Answer 1

At the bottom of the circle
From the FBD
R_N - mg = ma
where R_N is the normal reaction = 2.5(mg)
mg is weight of the motorcycle = 92*9.81
a is acceleration = V²/R
where R is the radius of circle = 15 m
Putting all the values , we get
2.5(mg) -mg = m(V²/R)
mg*(2.5-1) = m(V²/R)
V² = 1.5gR = 1.5*9.81*15
V = 14.857 m/s
Now the initial energy of the motorcycle = kinetic energy = (1/2)mV² = (1/2)*92*14.857² = 10153.6 J
Now when it reaches the side of the bottom
From the FBD
F_net = R_N
m*(v²/R) = 3mg
v² =3gR = 3*9.81*15
v = 21.01 m/s
Now final energy at the side = Potential energy + kinetic energy
= mgH + (1/2)mv²
= (92*9.81*15) +(1/2)*92*21.01² = 13537.8 + 20305.325 = 33843.125 J
Now the work done(W_net) = Final energy - initial energy
W_net = 33843.125 - 10153.6
W_net = 23689.525 J
hence this will be the net work done by all the forces.