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3 years ago

A student pushed a 100 N bicycle over a distance of 15 m in 5 s. calculate the power generated.

Physics

1 answer:

Umnica [9.8K]3 years ago

7 0

The catch in this one is: We don't know how much <u>force</u> the student used to push the bike.

It wasn't necessarily the 100N. That's just the weight of the bike. But you know that you can push a car, a wagon, or a bicycle hard, you can push it not so hard, you can give it a little push, you can give it a big push, you can push it strong, you can push it weak, you can push it medium. The harder you push, the more it'll accelerate, but it's completely up to you how hard you want to push. That's what's so great about wheels ! That's why they were such a great invention ! This is where I made my biggest mistake. This guy came into my store one day and said he's got this great invention, it's definitely going to take off, it'll be a winner for sure, he called it a "wheel". I looked at it, I turned it over and I looked on all sides. I thought it was too simple. I didn't know then it was elegant. I threw him out. I was so dumb. I could have invested money in that guy, today I would have probably more than a hundred dollars.

Anyway, can we figure out how much force the student used to push with ? Stay tuned:

-- The bike covered 15 meters in 5 seconds. Its average speed during the whole push was (15m/5s) = 3 meters/sec.

-- If the bike started out with no speed, and its average speed was 3 m/s, then it must have been moving at 6 m/s at the end of the push.

-- If its speed increased from zero to 6 m/s in 5 seconds, then its acceleration was (6m/s / 5 sec) = 1.2 m/s²

-- The bike's weight is 100N.

(mass) x (gravity) = 100N

Bikemass = (100N) / (9.8 m/s²)

Bikemass = 10.2 kilograms

-- F = m A

Force = (mass) x (acceleration)

Force = (10.2 kg) x (1.2 m/s²)

Force = 12.24 N

-- Work = (force) x (distance)

Work = (12.24 N) x (15 m)

Work = 183.67 Joules

-- Power = (work done) / (time to do the work)

Power = (183.67 joules) / (5 seconds)

<em>Power = 36.73 watts</em>

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