force and work question- what am I missing?

[samser]

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I'm going CRAZY I do NOT understand how EK came to this conclusion, maybe someone else has an easier, more logical method to getting to this final answer?

Essentially the question is about a picture of a box 'm' with a 10 N force pushing to the right ('N') and another force pushing to the left at an angle ('F' theta)...

So the question is Force F is 100 N, m is 10 kg and theta is 30 degrees. The coefficient of friction between the box and the surface is 0.1. If the block starts from rest, and is pushed for 5 m, what is the approximate final velocity of the block?
a- 0
b- 4.5 m/s
c- 5.8 m/s
d- 7.9 m/s

they say it's 7.9... does anyone have a logical way to get to this conclusion? I'm slamming my head on the table!

thanks!

 

[chiddler]

i keep getting something close to C!

:<

is it 30 degrees to the x or y axis / to the vertical or horizontal?

need mo' angle details

 

[jevo]

siht. I was wrong lol, nvm

 

[cloak25]

Hmm..I got C as well.

 

[Ineedhopenow]

I get a little over 8 with rounding, but not sure if I did it right.

10 N to the right plus the frictional force (0.1*100N= 10 N) equals to 20 N to the right.

100cos(30) = 90 N to the left. Therefore there's a net force of 70 N to the left.

I then placed work = kinetic energy so force (70 N) * distance (5 m) = 1/2 mass (10 kg) * velocity^2.

350 Nm = 1/2 *10*v^2
70 = v^2
~8.2 = v

 

[samser]

0

 

[chiddler]

that is very incorrect. normal force changes when you pull at an angle. the supposed solution above didn't do any accounting of that.

however seeing that solution made me realize i might have switched sin and cos

 

[MedPR]

We kind of need to know what direction the force is angled (up or down).

However, assuming the force is pointing downwards at 30deg to the horizontal, you can do it this way.

F=100cos30 - 10 - 100(0.1)
F=86-10-10
F=66
ma=66
a=6.6

x=1/2at^2
10=6.6t^2
t=sqrt(10/7)

a=v/t

10*sqrt(10/7)=v

sqrt10/7 is less than 2, so sqrt of 10/7 is less than 1.4 but greater than 1.

So 7*sqrt10/7 is greater than 7 but not by a lot.

Answer is D.

 

[Ineedhopenow]

that is very incorrect. normal force changes when you pull at an angle. the supposed solution above didn't do any accounting of that.

however seeing that solution made me realize i might have switched sin and cos

Is the box on a ramp? I thought it was on a flat surface.

 

[MedPR]

Is the box on a ramp? I thought it was on a flat surface.

It's on a flat surface, but if the force is angled upward, the normal force is reduced. Hence why if you pull a heavy object with a rope, you pull upward rather than pulling downward.

 

[411309]

We kind of need to know what direction the force is angled (up or down).

However, assuming the force is pointing downwards at 30deg to the horizontal, you can do it this way.

F=100cos30 - 10 - 100(0.1)
F=86-10-10
F=66
ma=66
a=6.6

x=1/2at^2
10=6.6t^2
t=sqrt(10/7)

a=v/t

10*sqrt(10/7)=v

sqrt10/7 is less than 2, so sqrt of 10/7 is less than 1.4 but greater than 1.

So 7*sqrt10/7 is greater than 7 but not by a lot.

Answer is D.

i also used this method.

 

[Ineedhopenow]

It's on a flat surface, but if the force is angled upward, the normal force is reduced. Hence why if you pull a heavy object with a rope, you pull upward rather than pulling downward.

Wow I didn't know that. So would the normal force be mgcos(theta)?

 

[typicalindian]

Wow I didn't know that. So would the normal force be mgcos(theta)?

if it's on a ramp, then yes. If it's on a flat surface then normal force is just mg

also,

We kind of need to know what direction the force is angled (up or down).

However, assuming the force is pointing downwards at 30deg to the horizontal, you can do it this way.

F=100cos30 - 10 - 100(0.1)
F=86-10-10
F=66
ma=66
a=6.6

x=1/2at^2
10=6.6t^2
t=sqrt(10/7)

a=v/t

10*sqrt(10/7)=v

sqrt10/7 is less than 2, so sqrt of 10/7 is less than 1.4 but greater than 1.

So 7*sqrt10/7 is greater than 7 but not by a lot.

Answer is D.

that seems like a lot more work than if you just use
work = KE

 

[MedPR]

if it's on a ramp, then yes. If it's on a flat surface then normal force is just mg

also,




that seems like a lot more work than if you just use
work = KE

Yea that's true.

v^2=66*5*2/10
v^2=66
v=8.1 or something. Close enough to D.

 

[Ineedhopenow]

It's on a flat surface, but if the force is angled upward, the normal force is reduced. Hence why if you pull a heavy object with a rope, you pull upward rather than pulling downward.

Typicalindian, do you also agree with MedPR's statement?

 

[typicalindian]

edit: wrong explanation

 

[Ineedhopenow]

When putting Work = KE, we're not accounting for the heat loss due to friction. That's why our answers are slightly above answer D. If I'm wrong, please correct me.

 

[MedPR]

When putting Work = KE, we're not accounting for the heat loss due to friction. That's why our answers are slightly above answer D. If I'm wrong, please correct me.

Maybe you didn't, but I did.

 

[typicalindian]

Maybe you didn't, but I did.

I don't see in your answer where you accounted for it?

 

[MedPR]

Technically the normal force is the same, there's just another force acting in the opposite direction so the net force in the Y direction is less (or more depending on how you look at it) However, even if it were pushing upward with that angle, I don't think it would need to be taken into account. I say this because in this case, even though the force is being applied at a downward angle, we don't add the Y component of the applied force to the normal force.

If you have a block sitting on the ground, normal force is mg.

If you push on the block at an upward angle, the normal force is now mg-Fsintheta because you are "helping" to keep the block from falling through the ground. Any upward force decreases the normal force. This is a similar idea to that of buoyant force. If an object is submerged, it still has its same weight (mg), but it appears less because the fluid or gas it is submerged in is helping to lift it up, thus reducing the normal force.

 

[MedPR]

I don't see in your answer where you accounted for it?

Force to the right is 10N
Variable force is 100N directed at a 30degree angle, so 100cos30.
Friction is normal*mu = 10.

Net force = 86-10-10 = 66.

Looking back at that guy's work, he included friction too. He calculated the applied force to be 90N, then net force to the left to be 70N. He took into account the opposing force as well as frictional force.

We are off bc we are ignoring the Y-component's contribution to friction. The frictional force is actually mu*(Normal+50)

86-10-15 = 61. sqrt61 = 7.8

By applying a force downward at an angle 30deg above the horizontal, we are effectively pushing the block down into the ground and thus increasing the frictional force

 

[typicalindian]

If you have a block sitting on the ground, normal force is mg.

If you push on the block at an upward angle, the normal force is now mg-Fsintheta because you are "helping" to keep the block from falling through the ground. Any upward force decreases the normal force. This is a similar idea to that of buoyant force. If an object is submerged, it still has its same weight (mg), but it appears less because the fluid or gas it is submerged in is helping to lift it up, thus reducing the normal force.

so then how come pushing at a downward angle doesn't increase the normal force? you're "helping" to push it in to the ground

edit: saw your edit in the above post medpr 😀

Force to the right is 10N
Variable force is 100N directed at a 30degree angle, so 100cos30.
Friction is normal*mu = 10.

Net force = 86-10-10 = 66.

Oh the force due to friction is equal to the energy lost to heat? riiight haha whoops

 

[MedPR]

so then how come pushing at a downward angle doesn't increase the normal force? you're "helping" to push it in to the ground

edit: saw your edit in the above post medpr 😀




Oh the force due to friction is equal to the energy lost to heat? riiight haha whoops

😀 I ignored it initially because I saw that the highest answer was, in fact, the correct answer. So I knew that everything would be ok as long as the calculation excluding the additional friction came out higher than the right answer (7.9, or whatever it is).

In other situations where the answers are closer together (and especially when we don't know the correct answer) the direction of the force does matter and you should certainly be as accurate as possible. I'm surprised this question writer didn't include an answer that was tailored especially for those who neglected to include the y-component of the force into the frictional force.