Question

Am I correct?

Answer 1

I agree with 1 of the three. I won't give answers, because I don't feel comfortable doing that. If you explain your reasoning, I can steer you in the right direction.

Answer 2

The picture had only three, but yes!
I won't be able to reply very much, but I'll try to!
So, for the first one, with the bicycle traveling uphill (upward angle slope), why did you pick \(5^\circ\)?

Answer 3

Yep! And I just know biking uphill is harder!! Your physics teacher would probably like to hear that the downhill component of gravitational force is greater, since the most vertical slope there is \(9^\circ\). So congrats!! We now agree.

And why'd you choose \(1.4^\circ\) for the next one?

Answer 4

And the 0.139 is \(\sin(8^\circ)\), which gives the downhill component of gravitational acceleration. So I agree!

And why do you think that the younger brother will win?

Answer 5

Haha, so you changed your mind! Anyway, I'd agree if it weren't a physics problem about coasting.
Really, the skilled bicyclist might choose the best path. Orrrr, he might throw the race to make his little bro happy. But we need to talk physics.

So! Gravitational acceleration is about \(9.8\rm\ m/s^2\). Then you consider the component of acceleration that is due to the downhill force only. The downhill force is something like
\(F_g\sin\theta\)
So the downhill acceleration is
\(\dfrac{F_g}m\sin\theta\)
which is
\(g\sin\theta\)

So gravitational acceleration is broken up just like the force.
Each bicyclist has the \(\it same\) acceleration then, if they have the same sloped hill. It's the same fraction of gravitational acceleration.

Sooo... same acceleration... Have you changed your mind?

Answer 6

Yep!

For the next three, I agree with two of them, and the one I disagree with is probably a misunderstanding. Since you seem to know what you're doing, I'll let you know which ones I agree with. If you have any questions, though, let me know. I'm just assuming you used the correct method to get to your result.

I agree with #4 and #6.
#4 uses \(F=ma\Rightarrow\dfrac Fm=a\) and then \(a=\dfrac{v_f-v_i}{\Delta t}\Rightarrow v_f = a\Delta t +v_i\)

#6 is splitting forces like you have been doing, but the trick was understanding the situation!! So you did well.

#5 is tricky. It asks for the \(\color{blue}{\it net}\) force on the cart. So, you might want to consider all the forces acting on it. But you don't know the man's force that he pushes with, unless you calculate it. But I have an easier way.
\(F_{net}=ma\).
You have the mass, and the acceleration is implied. The cart has a constant velocity. What does that mean for the acceleration?

Answer 7

Yep!! Acceleration is 0, and so force is 0. If there is no acceleration, a.k.a. change in velocity, then the net force is 0.

So! Any questions?

Answer 8

You're very welcome! Congrats! :)

Answer 9

Sorry to bump in but I believe the answer to (1) is 3.
My reasoning goes like this -

So if we want \[mgcos\] to be maximum we need cos theta max which is max at the smallest angle, ie 3.