# The force of a Powerful Motorcycle| Newton’s Second Law of Motion: Concept of a System| Dynamics| College Physics| Problem 4.10

### A powerful motorcycle can produce an acceleration of 3.50m/s² 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 force does the motorcycle exert backward on the ground to produce its acceleration if the mass of the motorcycle with the rider is 245 kg?

#### SOLUTION:

From Newton’s second law of motion, we know that

$net\:F=ma$

Considering the horizontal direction and let f be the resisting force, the equation becomes

$F-f=ma$

We are solving for F. Substituting the given values, we have

$F-400\:N=\left(245\:kg\right)\left(3.50\:m/s^2\right)$

$F=\left(245\:kg\right)\left(3.50\:m/s^2\right)+400\:N$

$F=857.5\:N+400\:N$

$F=1257.5\:N$