Need Help with a Worksheet? Get it Checked by Experts! [SOLVED]

[scjibran]

[SOLVED] Can someone check my Worksheet for me ...

Um for number 6 I also got 7kg and 15 lbs. Cuz 40 pounds sounds a bit much. Lemme know what you guys get ?thanks ...
- Jibran

Moderation note: scjibran requested the image of the worksheet be deleted from this post now that the assignment has been turned in and graded already because it showed a full name on the scanned image. Thanks to all who helped...scjibran did very well on the assignment.[/color]

 

[G01]

Hi again Jibran. You seem to be confused in number 5. The block WEIGHS 2300N. That is not its mass. You seem to be using the number 2300N incorrectly as the mass in that problem. Check your work there.

For number 6, you are not correct. HINT: Draw a force diagram and add up all the forces on the rope. Then see if you can use Newton's second Law to find the tension. In number 7, you are finding the acceleration incorrectly. Acceleration is not distance divided by time (check the units). Try to use the kinematic equations. HINT: List all the information you know about the motion of the block, starting position, ending position, initial velocity, etc. Then, choose a kinematic equation in which you can use that information to find the acceleration.

The rest of your work seems good. It seems to me that you understand the concepts pretty well, but you seem to have a little trouble using your knowledge of those concepts in quantitative problems. The best way to get better at this is to practice. So, try these problems again thinking about the hints I gave and get back to me if you get stuck!

 

[scjibran]

Hi again Jibran. You seem to be confused in number 5. The block WEIGHS 2300N. That is not its mass. You seem to be using the number 2300N incorrectly as the mass in that problem. Check your work there.

For number 6, you are not correct. HINT: Draw a force diagram and add up all the forces on the rope. Then see if you can use Newton's second Law to find the tension. In number 7, you are finding the acceleration incorrectly. Acceleration is not distance divided by time (check the units). Try to use the kinematic equations. HINT: List all the information you know about the motion of the block, starting position, ending position, initial velocity, etc. Then, choose a kinematic equation in which you can use that information to find the acceleration.

The rest of your work seems good. It seems to me that you understand the concepts pretty well, but you seem to have a little trouble using your knowledge of those concepts in quantitative problems. The best way to get better at this is to practice. So, try these problems again thinking about the hints I gave and get back to me if you get stuck!

OMg I didn't even look at weighs ...

Okay for number 5 the first one is still 2300 ...
For the second one I do 2300/9.8 = 234.693 x .65 = 152.55 = 150N
And the third part 2300/9.8 = 234.693 x (9.8+.65) = 2452.551 = 2500N

I think that's right ?

For number 6 I got 100 / (4.5+9.8) = 6.99 = 7N and X by 2.2 = 15.384 and because of sig figs I get 15 lbs

for the last one I got acceleration .80/.60/.60 = 2.25/.5 = 4.5 m/s^2
I think that's right ... I don't have a clue what to do after that ?

p.S. - Where I wrote Gravity and what not, those are suppose to be numbers ... I don't get how to get the numbers though. My friend got 5.4 and -12 but I don't know how she did it though ...

 

[Astronuc]

Be careful in writing out one's work so that it is clear, and be careful with units.

In #7, if the 1200 g block (1.2 kg) is falling vertically, the force of gravity is just the weight mg = 1.2 kg * 9.8 m/s² = ____________ N.

One good reference is Hyperphysics:

http://hyperphysics.phy-astr.gsu.edu/hbase/N2st (click on any image for the particular problem)

Equations of linear motion:
http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html

Equations of rotational motion compared with those of linear motion.
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

 

[Astronuc]

The problem of #6 is like the elevator problem.

There is a mass m which has a weight mg just due to gravity. Now if the mass is pulled upward with acceleration, one has to add that acceleration to that of gravity, so the tension on the cable or line is m(g+a).

If something is falling with an acceleration, a, then the effective weight is m(g-a), which is why we feel lighter when an elevator accelerates downward.

If one free falls, a = g and one's weight feels like m(g-g) = 0, i.e. weightless.

 

[scjibran]

Be careful in writing out one's work so that it is clear, and be careful with units.

In #7, if the 1200 g block (1.2 kg) is falling vertically, the force of gravity is just the weight mg = 1.2 kg * 9.8 m/s² = ____________ N.

One good reference is Hyperphysics:

http://hyperphysics.phy-astr.gsu.edu/hbase/N2st (click on any image for the particular problem)

Equations of linear motion:
http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html

Equations of rotational motion compared with those of linear motion.
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

okay 1.2 x 9.2 and since its falling its -12 ? right ?

 

[Astronuc]

okay 1.2 x 9.2 and since its falling its -12 ? right ?

Actually, 1.2 kg * 9.8 m/s² = 11.8 N, and yes if on takes acceleration or force as + going up, then in this case -11.8 N, which is just the weight.

 

[scjibran]

Actually, 1.2 kg * 9.8 m/s² = 11.8 N, and yes if on takes acceleration or force as + going up, then in this case -11.8 N, which is just the weight.

OMG For acceleration I use the distance forumula ... d = .5at^2

yea I got it ... thanks