Homework Help: Take Home Test (Solved, but Checking Answers) From AP Physics B (Dynamics)

1. Oct 11, 2008

trageserk

I got a take home test for AP Physics B and I've completed it. I just wanted to check with some people to see if they agree with some of the questions I had some difficulty with. I'll list the questions & how I thought they are solved. Any and all feedback is greatly appreciated- thanks for the help!

8. A student pulls a box of books on a smooth horizontal floor with a force of 100 N in a direction of 37 degrees above the horizontal. If the mass of the box and hte books is 40.0 kg, what is the acceleration of the box?
A) 0 m/s^2 B) 1.5 m/s^2 C) 1.9 m/s^2 D) 2.0 m/s^2 E) 3.3 m/s^2

I chose answer e- Sigma F(sub x)= F(sub t)- F(sub g in the x plane)= m (a)
I resolved F(sub g) into the x component 392*sin(37)=235.911 and plugged into the equation. I solved for a by doing 235.911 + 40(a) = 100. It didn't come out to exactly 3.3, but I figured with sig figs and what not it was the best choice.

13. Two boxes of masses m and 2m are in contact with each other on a frictionless surface. What is the acceleration of the more massive box? (In the diagram that is drawn there is a force F touching the box m which is touching box 2m)
A) acceleration due to gravity B)F/m C) F/(2m) D) F/(3M) E) F/(4M)

I chose D and figured that because they are touching, they must be accelerating at the same speed- I didn't think that the force could differentiate between two different boxes that are placed together and a single box equal to both masses added together.

20. A new planet is discovered that has twice the Earth's mass and twice the Earth's radius. On the surface of this new planet, a person who weighs 500 N on Earth would experience a gravitational force of
A) 125 N B)250 N C) 500 N D) 1000 N E)2000 N

We haven't covered this yet, so I have no clue what the answer is.

Free Response Question-
1979B2. A 10-kilogram block rests initially on a table as shown in cases I and II above. The coefficient of sliding friction between the block and the table is 0.2. The block is connected to a cord of negligible mass, which hangs over a massless frictionless pulley. In case I a force of 50 newtons is applied to the cord. In case II an object of mass 5 kilograms is hung on the bottom of the cord. Use g = 10 meters per second squared.

a. Calculate the acceleration of the 10-kilogram block in case I.

b. On the diagrams below, draw and label all the forces acting on each block in case II

c. Calculate the acceleration of the 10-kilogram block in case II.

Part A- I worked it out, said that SimgaF= F(sub tension) - F(friction) = 10 kg a
I think that because in part A, only a force is applied, it is equivalent to the tension force (50 N). I ended up getting 3 m/s^2 for a.

Part C- I gathered equations Sigma F(sub x)= F(tension) - F(friction) = m(sub 1)*(a)
and Sigma F(sub y) = m(sub 2) * gravity - F(tension)= M(sub 2) a
I solved for F(tension in the first equation and plugged that into the second and solved, and got an acceleration of 2 m/s^2

2. Oct 11, 2008

cepheid

Staff Emeritus
For problem 8, I get 1.997 m/s^2

The floor is *horizontal*. Gravity is completely counteracted by the normal force and the upward component of the applied pulling force. So it does not enter into the problem at all.

Edit: for clarity, resolving gravity into components when it is already entirely vertical makes no sense.

Last edited: Oct 11, 2008
3. Oct 11, 2008

cepheid

Staff Emeritus
Problem 13 looks okay to me (your reasoning, that is). If they are in contact, that force is essentially pushing their combined mass.

4. Oct 11, 2008

trageserk

That makes a lot more sense now. I wasn't really too sure with the original answer that I had chosen (question 8). Thanks for the quick response.

5. Oct 11, 2008

cepheid

Staff Emeritus
For number 20, look up Newton's Law of Gravitation. It determines the force due to gravity between two masses.

6. Oct 11, 2008

trageserk

Is it 250 N?

7. Oct 11, 2008

cepheid

Staff Emeritus
Yes, because the force is directly proportional to the mass (and so it doubles, if radius is constant), but inversely proportional to the square of the distance (and so it is quartered, if the mass is constant). The net effect is to halve the force.

8. Oct 11, 2008

trageserk

Thanks for the help.