How Does a Motorcycle Exert Force Backward to Accelerate?

AI Thread Summary
To determine the backward force exerted by a motorcycle to achieve a specific acceleration, the net force can be calculated using Newton's second law, Fnet = ma. For a 245 kg motorcycle accelerating at 3.50 m/s² with 400 N of resistance from friction and air drag, the total backward force is calculated as Fnet = (245 kg x 3.5 m/s²) + 400 N, resulting in 1257.5 N. The speed of the motorcycle does not directly impact the calculations, as the resistance forces are provided for that specific speed. However, it is noted that different speeds would alter the drag and torque characteristics in real-life scenarios. Understanding these dynamics is crucial for accurate force calculations in physics.
whereitsbeen
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Homework Statement



A 245 kg motorcyle & rider can produce an acceleration of 3.50 m/s sq while traveling 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?

Homework Equations



Force = mass x acceleration + Friction motion


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.
 
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whereitsbeen said:

Homework Statement



A 245 kg motorcyle & rider can produce an acceleration of 3.50 m/s sq while traveling 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?

Homework Equations



Force = mass x acceleration + Friction motion

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.
Your equation is wrong.

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.
 
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.
 
Thank You
 

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