Solve 6 Easy Homework Questions on Forces

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The discussion revolves around solving six homework questions related to forces and motion. Key points include the clarification that an object cannot move in a curved path without a force acting on it, and the net force on a stationary apple is zero, while it becomes -1 N when released due to gravity. In free fall, a 1-kg ball experiences a net force of -10 N, which is the weight of the ball. The action and reaction forces when kicking a football are equal, but the force exerted by the foot is greater than the force exerted on it by the ball. Lastly, the tug of war scenario illustrates that both participants will slide toward each other with equal acceleration on frictionless ice.
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Homework Statement


1.)is it possible to move in a curved path in the absence of a force? Defend your answer.
2.)What is the net force on a 1-N apple when you hold it at rest above your head? What is the net force when you release it?
3.)What is the net force acting on a 1-kg ball in free fall?
4.)When you kick a football, what action and reaction forces are involved, which force, if any, is greater?
5.)Does a baseball bat slow down when it hits a ball? Defend your answer.
6.) Two people of equal mass attempt a tug of war with a 12-m rope while standing on frictionless ice. When they pull of the rope, each of them slides toward the other. How do their accelerations compare, and how far does each person slide before they meet?


Homework Equations





The Attempt at a Solution



1.) I have no idea, I would think that the item would just follow the path, and if the path was curved it would curve too.
2.) 0, and then -10m/s because that's the rate at which it falls due to gravity.
3.)-10m/s?
4.)The foot kicking the ball, and the ball kicking the foot, and the foot kicking the ball would be greater.
5.) I don't know :(
6.)I don't know. :(


I have a test tommorow, and need answers before then, Thanks!
 
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Here are some hints.

1. Review Newton's 1st Law of Motion
2. Okay, but watch the units. They are asking for acceleration. Your answer is a velocity.
3. Again watch those units. They are asking for a force, not a velocity or acceleration. You may need to review Newton's 2nd Law of Motion.
4, 5, and 6. Review Newton's 3rd Law of Motion.
 
vandorin said:
2.) 0, and then -10m/s because that's the rate at which it falls due to gravity.
10m/s is acceleration due to gravity, the force would just be its weight
vandorin said:
3.)-10m/s?
Free fall implies that an object has reached terminal velocity. Now answer,what is terminal velocity?

vandorin said:
4.)The foot kicking the ball, and the ball kicking the foot, and the foot kicking the ball would be greater.
From what law did you get these ("The foot kicking the ball, and the ball kicking the foot") forces from?
why would you think and the foot kicking the ball would be greater this?
 
Hi vandorin! :smile:

I'll just add this to Redbelly98's good hints:
vandorin said:
1.) I have no idea, I would think that the item would just follow the path, and if the path was curved it would curve too.

I think you're misunderstanding "path". The question means "curved path" in the abstract sense … in other words, simply a curve. :smile:
rock.freak667 said:
Free fall implies that an object has reached terminal velocity.

I don't think so, and nor does wikipedia. "Free fall" basically means falling, with no strings attached etc! :smile:
 
tiny-tim said:
I don't think so, and nor does wikipedia. "Free fall" basically means falling, with no strings attached etc! :smile:
:confused: I always thought that at terminal velocity you are in free fall.
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

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