EK 1001 Phys 339 Work

[aln012]

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339. The diagram has a force of 10N pushing a mass of 10Kg to the left, and a (F) Force of 100N pushing the right side of the mass at a 60 degree from the horizontal. The coefficient of friction between the block and the surface is .1. If the block starts from rest, and is pushed for 5m, what is the approximate final velocity of the block?

A. 0m/s
B. 4.5m/s
C. 5.8m/s
D. 7.9m/s

The answer is D. I have no idea what their explanation is talking about; all I got was that the normal force is Fsin(theta), which I did wrong because I thought that the Normal force was just the force perpendicular to the object and its surface. Can anyone help me out? Thanks in advance!

 

[banner18]

OK this is a great problem! a little too mathy for mcat but they'll give you easier numbers on the real test.

big thing here is to a draw free body dia. and make fnet= fcause - foppose

so, ive set the force F as a "causing" force, aka movement to the left is positive.

break up F into x and y; we are gonna look at the x forces first. Fx=50sqrt3
howd i get this? cos30=x/100, and cos30= sq.rt. 3/2.

now using fnet we have so far: 50srt3-(10+frictionalforce)=ma

frictional force= uF(normal)
the **KEY** to remember is that the normal force (mg sintheta) is not just 100N (weight of the block), its going to be 100+ the y component of force F. so Fy= 100*sin30= 50.
So we got 50+100= 150N= normal force
.1*150=15N as the frictional force

now plug that back into the fnet.
50sqrt3-(10+15)=ma

sqrt.3 is about 1.7 (memorize this!!! sqrt 2= about 1.4) so appox. 50sqrt3=85
65=10*a
a=6.5

now plug into the "no time" projectile motion eq. vf^2=vi^2+2ad
vf^2= 65
vf= slightly higher than 8 but we did some appoximations so the real value will be lower than this.
7.9 is the answer.

hope that helped, also i hope i did this the right way..... if there is a faster way lemme know!

 

[aln012]

thanks a lot banner!

I'm still a little confused about adding the normal Force to get 150N

 

[MT Headed]

The normal force exists to keep the thing from crashing though the floor (or floating away). It pushes perpendicular (up) on your object so that the NET force vector does not have any component up (floating) or down (crashing).

The total downward force is 100(weight)+50(push) newtons. Therefore the normal force is also up, at 150 newtons.

Anything else and you either break the floorboards or your block will float away.

 

[aln012]

wow that makes a lot of sense. Thanks!