awesome, I was just making sure if I had right and it is!argo4 wrote:
that's right. Friction force=myu X normal force, so 72=myu X 411.6Sheen1101 wrote:
1) A 72 N force is needed to slide a 42-kg box across a flat surface at a constant velocity. What is the coefficient of kinetic friction between the box and the floor? (Use g = 9.8 m/s2).
okay I need help on this one. I worked everything out but I don't know if this answer is correct or not. First I did 42kg(9.8 m/s2) = 411.6. Then I did (72/411.6) = .1749 is this answer right or wrong?
I still have some troubles trying to figure these problems out.
1) An elevator cable can stand a maximum tension of 19000 N before breaking. The elevator has a mass of 485 kg, and a maximum acceleration of 2.19 m/s2. Engineering safety standards require that the cable tension never exceed two-thirds of the breaking tension. How many 65-kg people can the elevator safely accommodate? [Round down to the nearest integer.]
2) A skier starts from rest at the top of a 23° slope 1.0 km long. Neglecting friction, how long does it take to reach the bottom? (answer is seconds)
3) A police officer investigating an accident estimates from the damage done that a moving car hit a stationary car at 25 km/h. If the moving car left skid marks 41 m long, and if the coefficient of kinetic friction is 0.68, what was the initial speed of the moving car? (answer in km.h)
4) Two large crates, with masses 640 kg and 490 kg, are connected by a stiff, massless spring (k = 7.5 kN/m) and propelled along an essentially frictionless, level factory floor by a force applied horizontally to the more massive crate. If the spring compresses 4.9 cm from its equilibrium length, what is the applied force?
~ For this one I got -648.876 N by (7500)(-0.049)(490/640 + 1) but the answer is wrong and I do not know what I did wrong
I spent almost 2 hours trying to figure these problems but got nowhere, hoping you can help.
1) An elevator cable can stand a maximum tension of 19000 N before breaking. The elevator has a mass of 485 kg, and a maximum acceleration of 2.19 m/s2. Engineering safety standards require that the cable tension never exceed two-thirds of the breaking tension. How many 65-kg people can the elevator safely accommodate? [Round down to the nearest integer.]
2) A skier starts from rest at the top of a 23° slope 1.0 km long. Neglecting friction, how long does it take to reach the bottom? (answer is seconds)
3) A police officer investigating an accident estimates from the damage done that a moving car hit a stationary car at 25 km/h. If the moving car left skid marks 41 m long, and if the coefficient of kinetic friction is 0.68, what was the initial speed of the moving car? (answer in km.h)
4) Two large crates, with masses 640 kg and 490 kg, are connected by a stiff, massless spring (k = 7.5 kN/m) and propelled along an essentially frictionless, level factory floor by a force applied horizontally to the more massive crate. If the spring compresses 4.9 cm from its equilibrium length, what is the applied force?
~ For this one I got -648.876 N by (7500)(-0.049)(490/640 + 1) but the answer is wrong and I do not know what I did wrong
I spent almost 2 hours trying to figure these problems but got nowhere, hoping you can help.
2)mg sin(23)=ma
g sin(23)=a
d=.5at2
1000=.5 g sin(23) t2
t=22.8s
g sin(23)=a
d=.5at2
1000=.5 g sin(23) t2
t=22.8s
Last edited by argo4 (2009-02-02 16:18:08)
that's all you do? Dang I was over thinking that one too much. thanksargo4 wrote:
2)mg sin(23)=ma
g sin(23)=a
d=.5at2
1000=.5 g sin(23) t2
t=22.8s
Does anybody know how to do the remaining 3 problems? Especially on number 4 because I'm really frustrated with that one, can't seem to get it right.TheunforgivenII wrote:
I still have some troubles trying to figure these problems out.
1) An elevator cable can stand a maximum tension of 19000 N before breaking. The elevator has a mass of 485 kg, and a maximum acceleration of 2.19 m/s2. Engineering safety standards require that the cable tension never exceed two-thirds of the breaking tension. How many 65-kg people can the elevator safely accommodate? [Round down to the nearest integer.]
3) A police officer investigating an accident estimates from the damage done that a moving car hit a stationary car at 25 km/h. If the moving car left skid marks 41 m long, and if the coefficient of kinetic friction is 0.68, what was the initial speed of the moving car? (answer in km.h)
4) Two large crates, with masses 640 kg and 490 kg, are connected by a stiff, massless spring (k = 7.5 kN/m) and propelled along an essentially frictionless, level factory floor by a force applied horizontally to the more massive crate. If the spring compresses 4.9 cm from its equilibrium length, what is the applied force?
~ For this one I got -648.876 N by (7500)(-0.049)(490/640 + 1) but the answer is wrong and I do not know what I did wrong
I spent almost 2 hours trying to figure these problems but got nowhere, hoping you can help.
guess everybody is tired of physics......like meTheunforgivenII wrote:
Does anybody know how to do the remaining 3 problems? Especially on number 4 because I'm really frustrated with that one, can't seem to get it right.TheunforgivenII wrote:
I still have some troubles trying to figure these problems out.
1) An elevator cable can stand a maximum tension of 19000 N before breaking. The elevator has a mass of 485 kg, and a maximum acceleration of 2.19 m/s2. Engineering safety standards require that the cable tension never exceed two-thirds of the breaking tension. How many 65-kg people can the elevator safely accommodate? [Round down to the nearest integer.]
3) A police officer investigating an accident estimates from the damage done that a moving car hit a stationary car at 25 km/h. If the moving car left skid marks 41 m long, and if the coefficient of kinetic friction is 0.68, what was the initial speed of the moving car? (answer in km.h)
4) Two large crates, with masses 640 kg and 490 kg, are connected by a stiff, massless spring (k = 7.5 kN/m) and propelled along an essentially frictionless, level factory floor by a force applied horizontally to the more massive crate. If the spring compresses 4.9 cm from its equilibrium length, what is the applied force?
~ For this one I got -648.876 N by (7500)(-0.049)(490/640 + 1) but the answer is wrong and I do not know what I did wrong
I spent almost 2 hours trying to figure these problems but got nowhere, hoping you can help.
I think your FBD for that 4th question is wrong. You have the force F pulling the smaller crate.. the question says that it is applied to the more massive crate (ie. the bigger one). Also, it says the spring is compressed. In your diagram, the spring would stretch.
I would have drawn something like:
F = -k∆d, where k=7,500N/m, and ∆d = -0.049 (-ve as the spring is being compressed)
F = - 7500 x -0.049
F = 367.5N
Dunno about the rest of that question.
I would have drawn something like:
Umm, as for the force required to compress the spring 4.9cm:F -> (640kg crate) ~~~ (490kg crate)
Where ~~~ represents the spring
F = -k∆d, where k=7,500N/m, and ∆d = -0.049 (-ve as the spring is being compressed)
F = - 7500 x -0.049
F = 367.5N
Dunno about the rest of that question.
I've already tried 367.5 N for my answer and it was wrong, i'm really confused on this particular questionmcminty wrote:
I think your FBD for that 4th question is wrong. You have the force F pulling the smaller crate.. the question says that it is applied to the more massive crate (ie. the bigger one). Also, it says the spring is compressed. In your diagram, the spring would stretch.
I would have drawn something like:Umm, as for the force required to compress the spring 4.9cm:F -> (640kg crate) ~~~ (490kg crate)
Where ~~~ represents the spring
F = -k∆d, where k=7,500N/m, and ∆d = -0.049 (-ve as the spring is being compressed)
F = - 7500 x -0.049
F = 367.5N
Dunno about the rest of that question.
To do number (3) is it like this? At first 25,000/41 = 609.75 and then 609.75(0.68) = 414.63 km/h. Is that the right or no? My guess is no but I don't know what I did wrong.TheunforgivenII wrote:
guess everybody is tired of physics......like meTheunforgivenII wrote:
Does anybody know how to do the remaining 3 problems? Especially on number 4 because I'm really frustrated with that one, can't seem to get it right.TheunforgivenII wrote:
I still have some troubles trying to figure these problems out.
1) An elevator cable can stand a maximum tension of 19000 N before breaking. The elevator has a mass of 485 kg, and a maximum acceleration of 2.19 m/s2. Engineering safety standards require that the cable tension never exceed two-thirds of the breaking tension. How many 65-kg people can the elevator safely accommodate? [Round down to the nearest integer.]
3) A police officer investigating an accident estimates from the damage done that a moving car hit a stationary car at 25 km/h. If the moving car left skid marks 41 m long, and if the coefficient of kinetic friction is 0.68, what was the initial speed of the moving car? (answer in km.h)
4) Two large crates, with masses 640 kg and 490 kg, are connected by a stiff, massless spring (k = 7.5 kN/m) and propelled along an essentially frictionless, level factory floor by a force applied horizontally to the more massive crate. If the spring compresses 4.9 cm from its equilibrium length, what is the applied force?
~ For this one I got -648.876 N by (7500)(-0.049)(490/640 + 1) but the answer is wrong and I do not know what I did wrong
I spent almost 2 hours trying to figure these problems but got nowhere, hoping you can help.
let's see....
0.5mv2 + mgh = 0.5mv02 + mgh0 + WR
h and h0 = 0, so:
0.5mv2 = 0.5mv02 + WR
therefore:
0.5mv02 = 0.5mv2 - WR
WR = -Rs
R = umg where u = kinetic friction;
0.5mv02 = 0.5mv2 + umgs
remove all the m's and put in the values:
0.5v02 = 0.5*6.92 + 0.68*9.81*41
v0 = sqrt((0.5*6.92 + 0.68*9.81*41)/0.5)
v0 = 24.4 m/s = 88 km/h
oh wow I was way off....thanksJenspm wrote:
let's see....0.5mv2 + mgh = 0.5mv02 + mgh0 + WR
h and h0 = 0, so:
0.5mv2 = 0.5mv02 + WR
therefore:
0.5mv02 = 0.5mv2 - WR
WR = -Rs
R = umg where u = kinetic friction;
0.5mv02 = 0.5mv2 + umgs
remove all the m's and put in the values:
0.5v02 = 0.5*6.92 + 0.68*9.81*41
v0 = sqrt((0.5*6.92 + 0.68*9.81*41)/0.5)
v0 = 24.4 m/s = 88 km/h
I still want to know how to get this problem
1) Two large crates, with masses 640 kg and 490 kg, are connected by a stiff, massless spring (k = 7.5 kN/m) and propelled along an essentially frictionless, level factory floor by a force applied horizontally to the more massive crate. If the spring compresses 4.9 cm from its equilibrium length, what is the applied force?
~ For this one I got -648.876 N by (7500)(-0.049)(490/640 + 1) but the answer is wrong and I do not know what I did wrong
as well this problem
2) An elevator cable can stand a maximum tension of 19000 N before breaking. The elevator has a mass of 485 kg, and a maximum acceleration of 2.19 m/s2. Engineering safety standards require that the cable tension never exceed two-thirds of the breaking tension. How many 65-kg people can the elevator safely accommodate? [Round down to the nearest integer.]
1) Two large crates, with masses 640 kg and 490 kg, are connected by a stiff, massless spring (k = 7.5 kN/m) and propelled along an essentially frictionless, level factory floor by a force applied horizontally to the more massive crate. If the spring compresses 4.9 cm from its equilibrium length, what is the applied force?
~ For this one I got -648.876 N by (7500)(-0.049)(490/640 + 1) but the answer is wrong and I do not know what I did wrong
as well this problem
2) An elevator cable can stand a maximum tension of 19000 N before breaking. The elevator has a mass of 485 kg, and a maximum acceleration of 2.19 m/s2. Engineering safety standards require that the cable tension never exceed two-thirds of the breaking tension. How many 65-kg people can the elevator safely accommodate? [Round down to the nearest integer.]
A 43 kg skater rounds a 5.0 m radius turn at 6.3 m/s.
(a) What are the horizontal and vertical components of the force the ice exerts on her skate blades?
(b) At what angle can she lean without falling over?
on (a) I found the horizontal force but I can't find the vertical force
(a) What are the horizontal and vertical components of the force the ice exerts on her skate blades?
(b) At what angle can she lean without falling over?
on (a) I found the horizontal force but I can't find the vertical force
never mind I found the vertical force but now I do need help on this particular problem"Sheen1101 wrote:
A 43 kg skater rounds a 5.0 m radius turn at 6.3 m/s.
(a) What are the horizontal and vertical components of the force the ice exerts on her skate blades?
(b) At what angle can she lean without falling over?
on (a) I found the horizontal force but I can't find the vertical force
Suppose the angles shown in this figure below are 63° and 23°.
(a) If the left-hand mass is 2.1 kg, what should be the right-hand mass in order that it accelerates downslope at 0.64 m/s2?
(b) What should the right-hand mass be so that it accelerates upslope at 0.76 m/s2?
1) A core is wrapped around the rim of a flywheel .5m in radius, and a steady pull of 30N is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center. The moment of inertia of the wheel is 4 mN. Compute the angular acceleration of the wheel.
2) A bucket of water with a mass of 20 kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder .4m in diameter, also with a mass of 20 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20m to the water. Neglect the weight of the rope.
a) What is the tension in the rope while the bucket is falling?
b) With what speed does the bucket strike the water?
c) What is the time of fall?
d) While the bucket is falling, what is the force exerted on the cylinder by the axle?
3) A playground merry-go-round has a radius of 4.4m and a moment of intertia of 245 mN and turns with negligible friction about a vertical axle through its center.
a) A child applies a 25N force tangentially to the edge of the merry-go-round for 20 s. If the merry-go-round is initially at rest, what is its angular velocity after this 20 s interval?
b) How much work did the child do on the merry-go-round?
c) What is the average power supplied by the child?
they were checked today, it turns out i only didn't know how to do 2. the other two i found out how to do through friends.
2) A bucket of water with a mass of 20 kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder .4m in diameter, also with a mass of 20 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20m to the water. Neglect the weight of the rope.
a) What is the tension in the rope while the bucket is falling?
b) With what speed does the bucket strike the water?
c) What is the time of fall?
d) While the bucket is falling, what is the force exerted on the cylinder by the axle?
3) A playground merry-go-round has a radius of 4.4m and a moment of intertia of 245 mN and turns with negligible friction about a vertical axle through its center.
a) A child applies a 25N force tangentially to the edge of the merry-go-round for 20 s. If the merry-go-round is initially at rest, what is its angular velocity after this 20 s interval?
b) How much work did the child do on the merry-go-round?
c) What is the average power supplied by the child?
they were checked today, it turns out i only didn't know how to do 2. the other two i found out how to do through friends.
Last edited by haffeysucks (2009-02-05 12:35:20)
"people in ny have a general idea of how to drive. one of the pedals goes forward the other one prevents you from dying"
1) In the Figure, a constant external force P=110 N is applied to a 20-kg box, which is on a rough horizontal surface. The Force pushes the box a distance of 8.0 m, in a time interval of 4.0s, and the speed changes from v1=0.3 m/s to v2=2.6m/s. The work done by the external force P is closest to:
(a) 760 J
(b) 440 J
(c) 690 J
(d) 610 J
(e) 520 J
I need help with this problem and I don't quite understand it. There is also a hint on this homework and it is (Hint Use W(friction)+W(P)=K.Ef-K.Ei). I've tried that equation but didn't really work, that's why (b) and (d) are wrong. Help?
(a) 760 J
(b) 440 J
(c) 690 J
(d) 610 J
(e) 520 J
I need help with this problem and I don't quite understand it. There is also a hint on this homework and it is (Hint Use W(friction)+W(P)=K.Ef-K.Ei). I've tried that equation but didn't really work, that's why (b) and (d) are wrong. Help?
I say (a), cause:TheunforgivenII wrote:
1) In the Figure, a constant external force P=110 N is applied to a 20-kg box, which is on a rough horizontal surface. The Force pushes the box a distance of 8.0 m, in a time interval of 4.0s, and the speed changes from v1=0.3 m/s to v2=2.6m/s. The work done by the external force P is closest to:
http://www.webassign.net/userimages/83723?db=v4net
(a) 760 J
(b) 440 J
(c) 690 J
(d) 610 J
(e) 520 J
I need help with this problem and I don't quite understand it. There is also a hint on this homework and it is (Hint Use W(friction)+W(P)=K.Ef-K.Ei). I've tried that equation but didn't really work, that's why (b) and (d) are wrong. Help?
Work = Force x Displacement x CosØ
Annnnnnd thus
Work = 110 x 8 x Cos30
Work = 762J = 760J (two sig figs, as given in question data)
cool, thanks for helpmcminty wrote:
I say (a), cause:TheunforgivenII wrote:
1) In the Figure, a constant external force P=110 N is applied to a 20-kg box, which is on a rough horizontal surface. The Force pushes the box a distance of 8.0 m, in a time interval of 4.0s, and the speed changes from v1=0.3 m/s to v2=2.6m/s. The work done by the external force P is closest to:
http://www.webassign.net/userimages/83723?db=v4net
(a) 760 J
(b) 440 J
(c) 690 J
(d) 610 J
(e) 520 J
I need help with this problem and I don't quite understand it. There is also a hint on this homework and it is (Hint Use W(friction)+W(P)=K.Ef-K.Ei). I've tried that equation but didn't really work, that's why (b) and (d) are wrong. Help?
Work = Force x Displacement x CosØ
Annnnnnd thus
Work = 110 x 8 x Cos30
Work = 762J = 760J (two sig figs, as given in question data)
No worries...
It worked, rite?
It worked, rite?
yep it worked.mcminty wrote:
No worries...
It worked, rite?
I have another problem that I can't seem to get it right
At first I thought that part (b) would be the same for part (c) but no I was wrong, so now I'm stuck on how to get the last part.
~ I also have another question that I can't seem to get. (sorry if you cannot see the picture, you can right click and view image)
At first I thought that part (b) would be the same for part (c) but no I was wrong, so now I'm stuck on how to get the last part.
~ I also have another question that I can't seem to get. (sorry if you cannot see the picture, you can right click and view image)
Last edited by TheunforgivenII (2009-02-10 18:34:51)
for the first problem, to get instantaneous velocity take the derivative to get the velocity vector valued function. Then put in the time and that should get you your answer. I got 12.2 i + (11-4t)j, and t = 1.7s yields 12.2i and 4.2j.TheunforgivenII wrote:
I have another problem that I can't seem to get it right
http://img21.imageshack.us/img21/2744/98414856ej7.jpg
At first I thought that part (b) would be the same for part (c) but no I was wrong, so now I'm stuck on how to get the last part.
~ I also have another question that I can't seem to get. (sorry if you cannot see the picture, you can right click and view image)
http://img15.imageshack.us/img15/2931/55361812so3.jpg
Too lazy right now to figure out the second.
awesome thanks for the helpRandomSchl wrote:
for the first problem, to get instantaneous velocity take the derivative to get the velocity vector valued function. Then put in the time and that should get you your answer. I got 12.2 i + (11-4t)j, and t = 1.7s yields 12.2i and 4.2j.TheunforgivenII wrote:
I have another problem that I can't seem to get it right
http://img21.imageshack.us/img21/2744/98414856ej7.jpg
At first I thought that part (b) would be the same for part (c) but no I was wrong, so now I'm stuck on how to get the last part.
~ I also have another question that I can't seem to get. (sorry if you cannot see the picture, you can right click and view image)
http://img15.imageshack.us/img15/2931/55361812so3.jpg
Too lazy right now to figure out the second.
Standard "use the formula for uniform circular motion" question. Here's my crack at it anyway... (and yes, that was shot with my webcam.. >_>)TheunforgivenII wrote:
~ I also have another question that I can't seem to get. (sorry if you cannot see the picture, you can right click and view image)
To clarify, that is 1210 km/h = 336.11... m/s. And the answer I got was a minimum of 22.59km or 22.6km, depending on how many sig figs you use (I would use 3 cause of 1210km/h in given information).
EDIT: Umm... I think I'm wrong. The acceleration you get from a = v2/r is in m/s2, not "g"'s..
So... since 1g = 9.8m/s2, 5g would be 49m/s2 (or 49.5 if you are using g=9.81).
From here, just sub into a=v2/r, where a=49.5m/s2 (ill use this value, you can use the other if you want) and v=366.11... m/s.
r = (366.11)2/49.5
r = 2707.80m
Therefore r=2.71km (which sounds more reasonable)
so is this some crash course in physics or what?
you've gotten damn far in 1 month tbh.
you've gotten damn far in 1 month tbh.