- #1

h4x

- 4

- 1

Hey everyone, I have been out of High School for 2 years and just recently got into a Physics 111 class a month late! They have covered quite a bit of material and my memory from high school is hazy :( I've been going over some problems and a few simple ones are confusing me. Help with any of the questions is really appreciated. I swear I am not stupid !

1)

What I did was:

(Weight) = (Mass)(G)

(Weight)/(Mass) = (G)

40N/5kg = 8m/s^2

Then I took that an put it into this formula:

F(normal)-(Mass)(G) = (Mass)(Acceleration)

(46.2N)-(5kg)(8m/s^2)=5(Acceleration)

Acceleration= 1.24m/s^2

Im fairly certian this is wrong and I am also unsure about my second formula. Could anyone clear this question up for me, and if I used the incorrect formula, could you please explain why the one you use works? Thanks.

2)

I simply used F=MA. Turned it into F=(m-7)(a). Which then turned into

110N=(M-7)(2.5m/s^2) which I believe gave me 6.3KG which I believe is probably wrong. Again, I am sorry for being so stupid and I am sure this will be an easy one, but thanks :)

3) Ah, here is the difficult one (for me). I have completely forgotten high school trig so its making this very difficult. If someone could PLEASE explain how to do this one I will be eternally gratefull. I have a general idea on how to do it but my lack of knowledge in trig is really holding me back. The trig part of this question will benifit me the most. Here it is:

Now, I believe I have to find out how hard ONE child is pushing in the horizontal direction (x), double that, and then subtract the snows force. Then I simply add the mass of the child and the sled and put it into F=MA.

Im having trouble finding (x) because I am terrible at trig :( Thanks for any help guys :D

edit:

I found "x" to be 45.05 by making the thing a right triangle. Then I doubled it due to there being two kids pulling the sled at the same rate, so I now have 90.11N. I then subtracted the 37N of backwards force; leaving me with 33N. I then put it into F=MA and got 33N=(19+3.7)(A) which made A = 1.45m/s^2.

Thanks.

1)

**On a planet far away an astronaut picks up a rock. The rock has a mass of 5.00kg and on this particular planet; has a weight of 40.0N. If the astronaut exerts a force of 46.2N on the rock, what is its acceleration?**What I did was:

(Weight) = (Mass)(G)

(Weight)/(Mass) = (G)

40N/5kg = 8m/s^2

Then I took that an put it into this formula:

F(normal)-(Mass)(G) = (Mass)(Acceleration)

(46.2N)-(5kg)(8m/s^2)=5(Acceleration)

Acceleration= 1.24m/s^2

Im fairly certian this is wrong and I am also unsure about my second formula. Could anyone clear this question up for me, and if I used the incorrect formula, could you please explain why the one you use works? Thanks.

2)

**You are pulling your little sister on her sled over a frictionless surface. You exert a constant horz force of 110N. The sled has an acceleration of 2.5m/s^2. If the sled has a mass of 7kg, what is the mass of your sister?**I simply used F=MA. Turned it into F=(m-7)(a). Which then turned into

110N=(M-7)(2.5m/s^2) which I believe gave me 6.3KG which I believe is probably wrong. Again, I am sorry for being so stupid and I am sure this will be an easy one, but thanks :)

3) Ah, here is the difficult one (for me). I have completely forgotten high school trig so its making this very difficult. If someone could PLEASE explain how to do this one I will be eternally gratefull. I have a general idea on how to do it but my lack of knowledge in trig is really holding me back. The trig part of this question will benifit me the most. Here it is:

**To give a 19kg child a ride, two teenagers pull on a 3.7kg sled with ropes. Both teenagers pull with a force of 55N at an angle of 35degrees relative to the forward direction:**

..\

35\

----[Sled]

x

Hopefully you understand my poor diagram. In addition, the snow exerts a force of 57N in the opposite direction: Find the acceleration of the sled and child...\

35\

----[Sled]

x

Hopefully you understand my poor diagram. In addition, the snow exerts a force of 57N in the opposite direction: Find the acceleration of the sled and child.

Now, I believe I have to find out how hard ONE child is pushing in the horizontal direction (x), double that, and then subtract the snows force. Then I simply add the mass of the child and the sled and put it into F=MA.

Im having trouble finding (x) because I am terrible at trig :( Thanks for any help guys :D

edit:

**Ive done some more work on the 3rd question, if anyone could help verify what I did that would be great.**I found "x" to be 45.05 by making the thing a right triangle. Then I doubled it due to there being two kids pulling the sled at the same rate, so I now have 90.11N. I then subtracted the 37N of backwards force; leaving me with 33N. I then put it into F=MA and got 33N=(19+3.7)(A) which made A = 1.45m/s^2.

Thanks.

Last edited: