# Big Test Prep-Questions

1. Nov 30, 2004

### Tom McCurdy

Big Test Prep--Questions

Alright we are having a review test and I needed to go back to the stuff we did earlier this year and I am having some trouble remembering what to do....
:shy: how can i forget this stuff you ask... I don't know

7.
A 52.3 kg trunk is pushed 5.95 m at a constant speed up a 28.0 degree incline by a constant horizontal force. The coefficent of kinetic friction beeween the trunk and the incline is .19 Calculate the work done by (a) the applied force (i got b)

2. Nov 30, 2004

### James R

The work done by a force is

$$W = \vec{F}.\vec{s} = |\vec{F}||\vec{s}|cos \theta$$

where $\theta$ is the angle between the force vector and the displacement vector.

3. Nov 30, 2004

### Sirus

Just $$W=Fd\cos{\theta}$$. Can you work with the force vectors acting on the block and figure it out? Start with a FBD.

4. Nov 30, 2004

### Tom McCurdy

Indded that is the forumula, however it is in calculating F, that I am having difficulty
I did just realize that it is horizontal so the $$cos \alpha$$ will apply as opposed to if they were pushing directly at the 28 degree angle,

5. Nov 30, 2004

### Tom McCurdy

I am being really slow with friction, its been awhile I tried setting it up like follows

$$\sum{x=0}P-f cos\alpha=0$$
$$\sum{y=0}N-mp-fsin\alpha=0$$

however I have a strong feeling this is incorrect, I have also tried 2 other ways but they are more involved and require a lot more work to show, and I know they don't work

6. Nov 30, 2004

### Tom McCurdy

P, being the force
N- Normal
f= uk
there should also be spaces betwen 0 and p and 0 and N

7. Nov 30, 2004

### Parth Dave

The applied force is horizontal. So you have four forces: The normal force (points directly up), friction (pointing back), gravity (pointing downwards) and your applied force, pointing horizontally. These are the equations i get:

Fy = Fn - Asin28 - mgcos28 = 0
Fx = Acos28 - Ff - mgsin28 = 0

You might want to double check that. I may have made a mistake somewhere. Solve for normal force in Fy and then plug that into Fx. Then you can solve for your applied force. When I did it, i got a slightly different answer. That may have been an algebraic mistake (hopefully). Either way, do it for yourself and see what you come up with.

8. Nov 30, 2004

### Parth Dave

Yea, i made a small mistake. It does work.

9. Dec 1, 2004

### Tom McCurdy

thanks for you help it worked out when I did the work

10. Dec 1, 2004

### Tom McCurdy

$$Fy=N-Psin28-mgcos28 = 0$$

$$Fx =Pcos28 - f - mgsin28 = 0$$

$$N=Psin28-mgcos28$$

$$Pcos28-ukN-mgsin28 = 0$$

$$Pcos28-uk*(Psin28-mgcos28)-mgsin28=0$$

$$Pcos28-ukPsin28-ukmgcos28-mgsin28=0$$

$$P(cos28-uksin28)=ukmgcos28+mgsin28$$

$$P=\frac{ukmgcos28+mgsin28}{cos28-uksin28}$$

$$P=\frac{.19*52.3*9.8*cos28+52.3*9.8*sin28}{cos28*.19sin28}$$

$$P=411.474$$

$$W=P(dot)d$$

$$W=P*d*cos28$$

$$W=411.474*5.95*cos28=2161.69$$

thanks for the help, just wanted to show you that I actually used your advice

11. Dec 1, 2004

### Tom McCurdy

The thing I need help is with comming up with the initial values the
$$Fy=N-Psin28-mgcos28 = 0$$
$$Fy=N-Psin28-mgcos28 = 0$$

12. Dec 1, 2004

### Parth Dave

Well the only suggestion i can make is that you draw yourself a free-body diagram and then start labelling your forces. Resolve them into your x and y components. Then just add them up.