Need Help with Newtonian Physics Problems?

[wakejosh]

I have been working on my homework and I am not totally sure about a few of the problems. Please let me know if any of these are wrong, and If anyone can point me in the right direction I would really appreciate it. Thanks

4. Doug hits his hockey puck, giving it an initial velocity of 6.0 m/s. If the coefficient of kinetic friction between ice and puck is 0.05, how far will the puck slide before stopping?

- I used the equation d= v^2 / 2gu , giving me 6^2 / (2*9.82*0.05) and came up with an answer of; 37m

5. As a basketball player starts to jump for a rebound, he moves upward faster and faster until he leaves the floor. During this time that his feet (shoes) are in contact with the floor, the force of the floor on his shoes is:

a. bigger than his weight
b. equal in magnitude and opposite in direction to his weight
c. less than his weight
d. zero

- I am thinking its A; because there is extra force (more than just his weight) because he is pushing with his legs.

6. As a car skids with its wheels sliding on a road covered with ice and snow, the force of friction between the icy road and the tires will usually be:

a. greater than the weight of the car times the coefficient of the static friction
b. equal to the weight of the car times the coefficient of the static friction
c. less than the weight of the car times the coefficient of the static friction
d. greater than the weight of the car times the coefficient of the kinetic fricion

- I'm a little confused on this one, and I am not sure if it is C or D. I figure the frictional force should be less than the weight of the car, but I thought it should be kinetic and not static friction. So i can't really decide between the two, maybe because I interperated something wrong in my book.

7. A 3.0 kg bucket is being lowered by a rope into a 10m deep well, starting from the top. The tension in the rope is 9.8 N. The acceleration of the bucket will be:

a. 6.5 m/s/s downward
b. 9.8 m/w/w downward
c. zero
d. 3.3 m/s/s upward

I think its A; I got my answer by using; Net F/ m = a .giving me 19.66/ 3.0 kg = 6.5 m/s/s

 

[Astronuc]

Answers for 4, 5 and 7 are correct.

On 6, one must realize that the coefficient of kinetic friction is less than that of static friction (i.e. \mu_k\,<\mu_s). When the car slides, one applies kinetic friction, which would give less friction than if applied static friction, i.e. if one used the coefficient of static friction * weight of the car to calculate the friction.

 

[jjg242]

Answers for 4, 5 and 7 are correct.

On 6, one must realize that the coefficient of kinetic friction is less than that of static friction (i.e. \mu_k\,<\mu_s). When the car slides, one applies kinetic friction, which would give less friction than if applied static friction, i.e. if one used the coefficient of static friction * weight of the car to calculate the friction.

so the only possible answer is D?

 

[Astronuc]

so the only possible answer is D?

No.

The friction force of the car skiding on an icy road is given by the product of the coefficient of kinetic friction (appropriate for the icy road and tire) and the weight of the car. The friction force is certainly not greater than the weight of the car times the coefficient of the kinetic friction!

The coefficient of friction will change according to the condition of the tire and road. If the road or tire wet or oily, then the static and kinetic friction coefficients are less than if the road and tire are clean. Lubricants reduce friction.

When the tire is rotating, the surface of the tire is essentially fixed (static) with the road, so one can apply the static coefficient of friction to find the friction force (traction). When the tired stops rotating, but the vehicle is continuing, the tire is sliding (the vehicle is skidding), and one then applies the coefficient of kinetic friction to the problem.

 

[jjg242]

ok. I am getting confused now.

Originnaly I figured that Kinetic friction applied to this problem considering that the road is icy. But I thought the friction would be less than w*coefficient.

So are you saying you can use the static coeffient if we assume the tire is spinning without traction on the ice? so would it be C then?

 

[Astronuc]

Originnaly I figured that Kinetic friction applied to this problem considering that the road is icy.

The kinetic friction applies when two surfaces are sliding. Static friction applies only when surfaces are stationary.

In the case of a spinning wheel, with no traction, the wheel surface is moving relative to the road surface, so kinetic friction would apply.

In this case (car skidding), the friction force is given by \mu_k*mg (mg = weight of car) which is less than the static friction which is given by \mu_s*mg. So answer is C.

 

[jjg242]

thats what i figured, thanks so much for your help.

 

[cknox]

Bucket

Not too late to chime in, am I...

I think the answer to 7 would be C (zero). Whether lowered by hand or the rope going around a pulley, changing directions, would have the same tension, so we can ignore these mechanics.

Gravity is 9.8N, so a tension of 9.8N is counter acting gravity, resulting in 0 acceleration. If tension was more or less than 9.8N, then there would be acceleration in some direction. So, I think this answer would be C.