Question

10: A powerful motorcycle can produce an acceleration of $\boldsymbol{3.50\textbf{ m/s}^2}$ while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg?

The powerful motorcycle, a chopper, in Problem 10 of Ch. 4.3 develops force, such that it accelerates at 3.50 m/s2 at v = 90.0 km/h, as written. Familiarize yourself with the entire problem. TASKS: (1) Carefully construct a free body diagram of the chopper with horizontal AND vertical constituent forces acting on it at v = 90.0 km/h. (2) Calculate "the magnitude of the constituent force F the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg." Answer to the nearest 0.1 Newton of force. Note: the chopper exerts a force backward on the ground; Newton's third law says that the ground exerts the same size force but forward, on the chopper.

Answer 1

Given,

a = 3.5 m/s^2 ; v = 92 kmph ; F = 400 N ; m = 245 kg

The amount of force that it exerts backwards will be:

Fb = ma + F(friction)

Fb = 245 x 3.5 + 400 = 1257.5 N

Hence, Fb = 1257.5 N