# Homework Help: Newtons Law - backward motion

1. Feb 22, 2012

### whereitsbeen

1. The problem statement, all variables and given/known data

A 245 kg motorcyle & rider can produce an acceleration of 3.50 m/s sq while travelling at 25 m/s. At that speed, the forces resisting motion (friction & air) total 400 N. What force does the motorcycle exert backward to produce it's acceleration?

2. Relevant equations

Force = mass x acceleration + Friction motion

3. The attempt at a solution

Fnet = (m x a) + Ffr
= (245 x 3.5) + 400N
= 1257.5N

This needs a 2nd look. I'm not sure if the speed matters since we have mass and acceleration values. Please take a look, I'll stand by.

2. Feb 22, 2012

### cepheid

Staff Emeritus
Welcome to PF!

Fnet = ma

Always. That is Newton's 2nd Law.

So, the question is, what does Fnet consist of? It is the sum of all horizontal forces on the motorcycle. There's the force from the road on the tires, which is forwards, and then there's friction + air drag, which is backwards.

These two have to add up to the net force, which you already know.

3. Feb 22, 2012

### cepheid

Staff Emeritus
The speed "matters" to the real-life situation in the sense that, at a different speed, the amount of drag would be different (and amount of torque the engine could produce would be different).

However, it doesn't matter to you, since it doesn't affect the actual calculations in any way. You are given what the drag + friction is at this speed.

4. Feb 22, 2012

Thank You