Couple Multiple Choice questions on force and centripetal acceleration.

In summary, the maximum speed at which the stone may be whirled in order for the string to not break is approximately 10 m/s. This is calculated using the equation F=mv2/r and solving for v, with a mass of .25 kg, a radius of 75cm, and a tension of 300N.The gravitational force on the moon due to the Sun is twice the gravitational force on the Moon due to the Earth. However, the moon is not pulled away from the Earth because both objects are in a stable orbit around the Sun, making the net force on the moon zero.When a car tows a trailer up a hill, the car exerts a greater force on the trailer than the trailer ex
  • #1
DavidDishere
12
0
#1) A child ties a .25kg rock to the end of a string and whirls it at a constant speed in a horizontal circle of radius 75 cm. What, approximately, is the maximum speed at which the stone may be whirled if the string will break for a tension exceeding 300N?

A. 10 m/s
B. 13 m/s
C. 15 m/s
D. 20 m/s
E. 30 m/s

I'm getting like 51 on this one so I must be doing something wrong.

#2) The gravitational force on the moon due to the Sun is about twice the gravitational force on the Moon due to the Earth. The Moon isn't pulled away from the Earth by the Sun because

A. it is much closer to the Earth than to the Sun.
B. the net force on it is zero.
C. both the Moon and the Earth accelerate together around the Sun.
D. all of the above
E. None of the above.

A and C both sound right. Is the net force on it zero though? I guess so right? So C?

#3) A car towing a trailer accelerates up a hill. Which one of the following statements is true?

A. The car exerts a smaller force on the trailer than the trailer exerts on the car.
B. The car exerts the same force on the trailer as the trailer exerts on the car.
C. The car exerts a greater force on the trailer than the trailer exerts on the car.
D. The relationship between the force exerted by the car on the trailer and the force exerted by the trailer on the car depends on the acceleration of the trailer.
E. The relationship between the force exerted by the car on the trailer and the force exerted by the trailer on the car depends on the mass of the trailer.

B. They are equal right?

#4) The coefficients of kinetic and static friction between a 3.0 kg crate and the florr on which it rests are .30 and .40 respectively. If the crate is initially at rest, what is the magnitude of the net force on the crate when a 6.0 N horizontal force is applied?

A. 0.0 N
B. 2.8 N
C. 5.8 N
D. 6.0 N
E. 17 N

It's A right? Because the normal force on the box is 29.4 N and then multiplied by static friction is 11.76 N so you would need to exert more than that to move the box. So if it's not moving, the net force is 0 right?
 
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  • #2
Hello! I believe I'm right with the below, of course. If I've made a mistake, I'm sure someone will chime in and correct me.

DavidDishere said:
#1) A child ties a .25kg rock to the end of a string and whirls it at a constant speed in a horizontal circle of radius 75 cm. What, approximately, is the maximum speed at which the stone may be whirled if the string will break for a tension exceeding 300N?
...
I'm getting like 51 on this one so I must be doing something wrong.

Can you show what equation(s) you're using for this one, and how you plugged the numbers in?

DavidDishere said:
#2) The gravitational force on the moon due to the Sun is about twice the gravitational force on the Moon due to the Earth. The Moon isn't pulled away from the Earth by the Sun because

A. it is much closer to the Earth than to the Sun.
B. the net force on it is zero.
C. both the Moon and the Earth accelerate together around the Sun.
D. all of the above
E. None of the above.

A and C both sound right. Is the net force on it zero though? I guess so right? So C?

Both the Earth and the moon are in free fall around the sun. The moon isn't pulled away because both it and the Earth are in a stable orbit around the sun. Yes, C is correct.

DavidDishere said:
#3) A car towing a trailer accelerates up a hill. Which one of the following statements is true?

A. The car exerts a smaller force on the trailer than the trailer exerts on the car.
B. The car exerts the same force on the trailer as the trailer exerts on the car.
C. The car exerts a greater force on the trailer than the trailer exerts on the car.
D. The relationship between the force exerted by the car on the trailer and the force exerted by the trailer on the car depends on the acceleration of the trailer.
E. The relationship between the force exerted by the car on the trailer and the force exerted by the trailer on the car depends on the mass of the trailer.

B. They are equal right?

Incorrect. If they were equal, the net force on the car-trailer system would be zero, and neither would accelerate.

DavidDishere said:
#4) The coefficients of kinetic and static friction between a 3.0 kg crate and the florr on which it rests are .30 and .40 respectively. If the crate is initially at rest, what is the magnitude of the net force on the crate when a 6.0 N horizontal force is applied?

A. 0.0 N
B. 2.8 N
C. 5.8 N
D. 6.0 N
E. 17 N

It's A right? Because the normal force on the box is 29.4 N and then multiplied by static friction is 11.76 N so you would need to exert more than that to move the box. So if it's not moving, the net force is 0 right?

I didn't check your math, but assuming your math is correct, your answer is correct. However, your last sentence there should read, 'So if it's not accelerating, the net force is 0.

Hope this helps!
 
  • #3
I just did 300 = mv2/r and solved for v. I got 51. I guess if I divide that by 2*pi*r, I get about 10.8 so maybe that is what I'm missing.
 
  • #4
DavidDishere said:
I just did 300 = mv2/r and solved for v. I got 51. I guess if I divide that by 2*pi*r, I get about 10.8 so maybe that is what I'm missing.

No, not the 2*Pi*r thing. I solved the equation and got something different, so you've made a mistake with either the algebra or your units.

[tex]F = \frac{mv^2}{r} => v = \sqrt{\frac{Fr}{m}}[/tex]
 
  • #5
Ahhh I was squaring m and r for some reason. Thanks.

So C is the only correct answer for 2? I meant to say that I was thinking it was C but then could see that A and B might be right too. So maybe it was D.

And can anyone explain the trailer problem? I guess I'm thinking of the car and trailer being one unit when they are hitched.
 
  • #6
DavidDishere said:
So C is the only correct answer for 2? I meant to say that I was thinking it was C but then could see that A and B might be right too. So maybe it was D.

I look at it like this. I believe C is the correct answer. At this point, C is my strongest candidate, and I've ruled out E, none of the above. This leaves, A, B, and D. If A and B are both also right, then D (all of the above) is the correct answer. If A and B are both wrong, then C is the correct answer. If only A or B (and C) are correct, well... that can't be, because that's not an option on this problem. So from a question-answering standpoint, these are the options.

Is A correct? It is true that the moon is much closer to the earth. Is this why the moon isn't pulled away from the earth, compared to the sun? Logically, it can't be, since despite the distances, the force on the moon due to the sun is, according to the problem statement, "about twice the gravitational force on the Moon due to the Earth." So I've pretty much ruled this out.

At this point, I can safely answer C. However, it would be nice, from a question-answering standpoint, if B were also incorrect. Kind of reinforces that C is the only answer. Is it? B says, "the net force on the moon is zero." Well, the problem statement itself says otherwise, as quoted in the previous paragraph-- the net force is non-zero, in the direction of the sun. So B is incorrect.

This leaves C.

DavidDishere said:
And can anyone explain the trailer problem? I guess I'm thinking of the car and trailer being one unit when they are hitched.

I'll give it a go, although hopefully someone else will jump in too and give another viewpoint, it often helps to hear things different ways since we all think a little differently.

The way I look at it is this; imagine the car and the trailer are on the hill, at rest. What forces are acting on them at this point? Well, the car is being pulled somewhat down the hill by gravity, and so is the trailer. If there were no brakes being used, the car and the trailer would begin rolling (accelerating) down the hill. If they are attached, they will move together.

Now imagine the car and the trailer are attached, and the car has its parking brake engaged. What forces are they experiencing here? The same as before, gravity pulling each down the hill. Additionally, though, the brake is applying an opposite force, preventing them from rolling down the hill. But the trailer doesn't have any brakes. It just hangs from the car. You can imagine, then, the trailer applying a force on the car, trying to pull it backwards. The car's brake has to work even harder to keep the vehicle stationary with a trailer attached.

Now imagine the car accelerating up the hill, like in the problem statement. The trailer doesn't have a motor, so it can't accelerate itself up the hill; it must be pulled by the car. The trailer is still applying this force to the car, trying to drag it back down the hill. But the car's engine is applying a greater force up the hill, accelerating them both up the hill.

It would help if you had ever towed a trailer up a hill before, or anything else. The engine moans and groans, and you accelerate really slowly. The engine has to work that much harder to get the whole rig moving. Why? Because the trailer is heavy, and it wants to pull the car down the hill.

Hope this helps!
 
  • #7
I understand what you are saying for #3 but what does that make the answer? C?
 

1. What is the difference between force and centripetal acceleration?

Force is a physical quantity that causes a change in an object's motion or shape, while centripetal acceleration is the acceleration experienced by an object moving in a circular path.

2. How are force and centripetal acceleration related?

Force and centripetal acceleration are related because force is what causes an object to experience centripetal acceleration. In other words, force is responsible for keeping an object moving in a circular path.

3. What is the formula for calculating centripetal acceleration?

The formula for calculating centripetal acceleration is a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular path.

4. How does centripetal acceleration affect an object's motion?

Centripetal acceleration causes an object to continuously change direction, but not speed. This means that the object is constantly accelerating towards the center of the circular path it is moving in.

5. Can centripetal acceleration change the speed of an object?

No, centripetal acceleration does not change the speed of an object. It only changes the direction of the object's motion, keeping it moving in a circular path at a constant speed.

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