Stormer's Question Thread

[superflya]

hey mao, pm'ed u yesterday. did u get it?

[QuantumJG]

Ok so let's look at the question carefully.

ΔW = ΔK => 0.5mΔv² = F x Δs

let m = 0.1kg => Δv = sqrt(240)m/s (~56m/s - reasonable)

let Δs = 1m (I would say an overarm throw is ~1m) => F = 12N

So the other parts are really redundant.

          

[Stormer]

What do you mean by the s? Is that the length of the arm or is that how far the ball has been thrown?

[QuantumJG]

What do you mean by the s? Is that the length of the arm or is that how far the ball has been thrown?

It's how far your arm has moved.

[Stormer]

Can you just go ΔK = 12J

So 12J = F x ΔS?

Gives the answer of 12 still. I don't see why it's particular necessary to estimate the mass. This would mean that only one estimation is needed as well.

[Akirus]

I wouldn't worry about it too much, usually they give you all of the figures to use (at least, that's been the case in every question I've ever seen or done).

[Mao]

Can you just go ΔK = 12J

So 12J = F x ΔS?

Gives the answer of 12 still. I don't see why it's particular necessary to estimate the mass. This would mean that only one estimation is needed as well.

exactly

[Akirus]

Even if you use W=Fx, you're still making an assumption (that the throw is made in 0.13s) so your answer has similar accuracy either way. For example, if the throw were actually made in 0.2 or 0.5 seconds, the average horizontal force applied would be lower.