Answer :
The centripetal force on the motorcycle at the top of the hill is 1696.42 N and the normal force is 1723.57 N.
The speed of the motorcycle is 25 m/s and the mass of the motorcycle is 342 kg and the radius of curvature of the hill is 126m.
(a) The centripetal force on any curvature is given by,
F =MV²/R
Where,
F is the centripetal force,
M is the mass of the motorcycle,
R is the radius of curvature of the hill,
V is the speed at which it is travelling.
Putting all the values,
F = 342 x 25 x 25/126
F = 1696.42 N.
The centripetal force is 1696.42 N.
(b) At the top of the hill, the centripetal force and the normal force will be away from the ground and in opposite direction of the weight of the motorcycle,
N = W- F
W is the weight,
F is centripetal force and N is the normal force.
Putting values,
N = 3420 - 1696.42
N = 1723.57N.
The normal force is 1723.57 N.
To know more about Centripetal force, visit,
https://brainly.com/question/20905151
#SPJ4
