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Old Oct 22nd 2012, 12:39 PM   #1
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Physics Worksheet Help(7 problems)

I need all forces required to complete a free body diagram as well as an explaination so I would be able to complete another problem like it. As far as my attempt. On most of them, I have no idea where to even being except for 1,3, and 6. I will post my attempts beneath those problems. Even if you don't want to do all of them, any help would be appreciated.
1) A 1.2*10^3 kg car is accelerating at 1.6 m/s. If the coefficient of friction is .15, what is the force supplied by the engine?

Attempt: I started out with my free body diagram. I got my weight, -11760 by multiplying my mass of the car by the force of gravity -9.8. I got the normal force by taking away the nagative sign. I didn't know my force, nor the force of friction so I left those blank. I did my x y charts, and my y's added up to 0, and I set my Ffx to ma. But I'm not sure what to do from there
2) A car is travelling at 120 km/hr when it slams on the brakes. How long is the skid mark if the coefficient of friction is .62?

3)A skidder is dragiing a 5.2*10^2 kg log through the forest at a constant speed of 3.5 m/s. If the skidder is applying a force of 1.8*10^3N to the log to keep it moving, what is the coefficient of friction between the log and the ground?

Attempt: I simply started out saying that my normal force was 5200, by gravity was -5200, and my force was 1800N. Then knowing that my force was 1800 I set up my equation so that it looked like -1800=520*uk. I then got my uk equals -3.46. I don't know how it is possible to be negative, nor do I think the value would be right if it was positive.

4)A curler gives a rock an initial velocity of 4.2 m/s. After travelling down the 32m ice sheet(coefficient of kinetic friction=.0035) the rock runs onto the carpet(ce of friction=.41). How far does the rock slide on the carpet?

5)A 2.0*10^3 kg truck is driving down the road at 105km/hr when a dog runs out on the road 94m in front of it. The driver's reaction time is .35 seconds before hitting the brakes and the coefficient of friction between the tires and the road is .85. Will the truck hit the dog?

6)A tow truck is trying to pull a 1.4*10^3 kg car out of some mud. The coefficient of static friction is .76. What force will the two truck have to apply to the car before it will start to move?

Attempt: For this one I said that my weight was -1400(9.8) and my normal force was +1400(9.8). I didn't know my force though. I set up my equation so that F=(.76)*(1400*9.8) and I got 10427N, which I think is way too much, so I believe I messed up somewhere, but I have no idea where.

7)A 3.2*10^3 kg sailboat is sailing at 6.2 knots(1kt=1.852km/hr) when the wind does. The boat drifts for 65m before coming to a stop.
a) What is the coefficient of friction between the hull and the water?
b)How long does the boat take to stop?
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Old Oct 22nd 2012, 02:52 PM   #2
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I'll give you some hints:

Originally Posted by deathring109 View Post
1) A 1.2*10^3 kg car is accelerating at 1.6 m/s. If the coefficient of friction is .15, what is the force supplied by the engine?
The force supplied by the engine must equal the force required to accelerate the car plus the force required to overcome friction. So F = ma + Wu, where W is the car's weight (W=mg) and u is the coefficient of friction.

Originally Posted by deathring109 View Post
2) A car is travelling at 120 km/hr when it slams on the brakes. How long is the skid mark if the coefficient of friction is .62?
Use energy principals - the initial kinetic energy of the car must equal the force of the brakes times the distance over which that force is applied. The force of the brakes is F=Wu.

Originally Posted by deathring109 View Post
3)A skidder is dragiing a 5.2*10^2 kg log through the forest at a constant speed of 3.5 m/s. If the skidder is applying a force of 1.8*10^3N to the log to keep it moving, what is the coefficient of friction between the log and the ground?

Attempt: I simply started out saying that my normal force was 5200, by gravity was -5200, and my force was 1800N. Then knowing that my force was 1800 I set up my equation so that it looked like -1800=520*uk.
I then got my uk equals -3.46. I don't know how it is possible to be negative, nor do I think the value would be right if it was positive.
....
Instead of 520 you should use the skidder's weight, which is W=mg. You also don't need the minus sign, since the friction force is weight times coefficient of friction and is a positive number opposing the skidder's direction of travel. The coeffeicient of friction is always a positive number, and is almost always a value less than 1.

Originally Posted by deathring109 View Post
4)A curler gives a rock an initial velocity of 4.2 m/s. After travelling down the 32m ice sheet(coefficient of kinetic friction=.0035) the rock runs onto the carpet(ce of friction=.41). How far does the rock slide on the carpet?
Again, use energy principals. Find the velocity at the end of the ice using delta KE = friction force times distance, then use that same equation for the carpet to determine distance travelled along the carpet to the point where v=0.

Originally Posted by deathring109 View Post
5)A 2.0*10^3 kg truck is driving down the road at 105km/hr when a dog runs out on the road 94m in front of it. The driver's reaction time is .35 seconds before hitting the brakes and the coefficient of friction between the tires and the road is .85. Will the truck hit the dog?
Find the distance travelled under braking by again using the energy equation: Delta KE = friction force times distance. Again, the friction force is Wu.

Originally Posted by deathring109 View Post
6)A tow truck is trying to pull a 1.4*10^3 kg car out of some mud. The coefficient of static friction is .76. What force will the two truck have to apply to the car before it will start to move

Attempt: For this one I said that my weight was -1400(9.8) and my normal force was +1400(9.8). I didn't know my force though. I set up my equation so that F=(.76)*(1400*9.8) and I got 10427N, which I think is way too much, so I believe I messed up somewhere, but I have no idea where.
Actually, I think you are correct on thsi one!

Originally Posted by deathring109 View Post
7)A 3.2*10^3 kg sailboat is sailing at 6.2 knots(1kt=1.852km/hr) when the wind does. The boat drifts for 65m before coming to a stop.
a) What is the coefficient of friction between the hull and the water?
As before - use the energy equation.

Originally Posted by deathring109 View Post
b)How long does the boat take to stop?
Calculate the deceleration of the boat from F=ma. Then the distance travelled under that deceleration is d = 1/2 at^2. You were given the value for 'd', so solve for 't.'
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Old Oct 22nd 2012, 09:48 PM   #3
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Originally Posted by ChipB View Post
I'll give you some hints:



The force supplied by the engine must equal the force required to accelerate the car plus the force required to overcome friction. So F = ma + Wu, where W is the car's weight (W=mg) and u is the coefficient of friction.



Use energy principals - the initial kinetic energy of the car must equal the force of the brakes times the distance over which that force is applied. The force of the brakes is F=Wu.



Instead of 520 you should use the skidder's weight, which is W=mg. You also don't need the minus sign, since the friction force is weight times coefficient of friction and is a positive number opposing the skidder's direction of travel. The coeffeicient of friction is always a positive number, and is almost always a value less than 1.



Again, use energy principals. Find the velocity at the end of the ice using delta KE = friction force times distance, then use that same equation for the carpet to determine distance travelled along the carpet to the point where v=0.



Find the distance travelled under braking by again using the energy equation: Delta KE = friction force times distance. Again, the friction force is Wu.



Actually, I think you are correct on thsi one!



As before - use the energy equation.



Calculate the deceleration of the boat from F=ma. Then the distance travelled under that deceleration is d = 1/2 at^2. You were given the value for 'd', so solve for 't.'
I was only able to use your hellp for number 1. We haven't done energy principals or anything like that yet. We've only done kinematics, friction/non friction and vectors
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Old Oct 23rd 2012, 04:33 AM   #4
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OK, instead of energy principals you can use this equation for many of these problems:



The acceleration 'a' can be found from F=ma, where 'F' is the friction force Wu. So this resolves to a = ug where 'u' is the coefficient of friction. Now given v1, v2, and u, you can solve for d.
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Old Oct 23rd 2012, 07:15 PM   #5
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Originally Posted by ChipB View Post
OK, instead of energy principals you can use this equation for many of these problems:



The acceleration 'a' can be found from F=ma, where 'F' is the friction force Wu. So this resolves to a = ug where 'u' is the coefficient of friction. Now given v1, v2, and u, you can solve for d.
I talked to my physics teacher today and she helped me out with most of the problems. I did all of them, but #7 I feel I did wrong.

I started of converting my speed to m/s, so from 6.2 knots, I got 3.19 m/s

I then found acceleration by using the equation vf2=vi2+2a(x-xi) to get a=.008
for a I set up my equation so that u*(3200*9.8)=3200*.08. I solved and got u=.008 which was my answer to question a.

for b I set up a kinematics table so that...
xi-0
xf-65m
vi-3.19m/s
vf-0
a-.008
t-?

I then used the equation x=xo+.5(vo+vf)t(since I didn't want to have to use acceleration because I felt I calculated it wrong earlier) and plugged in my varibles to get 40.8 seconds.

Both answers seem resonable, but I have a feeling like a made a math error somewhere. I was able to solve all of the other ones fairly easy, but I don't know about this one
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Old Oct 24th 2012, 01:04 AM   #6
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I didn't find any mistake in there

Good job!
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Old Oct 25th 2012, 05:10 AM   #7
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waow! I read it and i like it. This is very helpful for me. Thanks for sharing the awesome post.
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