Any help pleeeze!! I'm totally lost on these!!
1. A 14 g coin slides upward on a surface that is inclined at an angle of 16° above the horizontal. The coefficient of kinetic friction between the coin and the surface is 0.23; the coefficient of static friction is 0.31. Find the magnitude and direction of the force of friction under the following circumstances.
(a) when the coin is sliding
.03 N
parallel to the incline
up the incline
down the incline
(b) after it comes to rest
.012 N
parallel to the incline
down the incline
up the incline
2. A 80 kg astronaut and a 1300 kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving it a speed of 0.23 m/s directly away from the shuttle. Seven-and-a-half seconds later the astronaut comes into contact with the shuttle. What was the initial distance from the shuttle to the astronaut?
1.725 m
3. Find the orbital speed of a satellite in a circular orbit 2540 km above the surface of the Earth.
215320 km/s
4. The force shown in Figure 7-16 acts on a 1.7 kg object whose initial speed is 0.44 m/s and initial position is x = 0.27 m.
Figure 7-16
(a) Find the speed of the object when it is at the location x = 0.85 m.
.74 m/s
(b) At what location would the object's speed be 0.40 m/s?
.46 m
one question per post please.
On the first, how is static friction less than sliding friction?
On the second, I don't get that answer. I will be happy to critique your work.
I didn't look at three or four (no figure).
I'm having problems with a friction question. here it is..
1. A 14 g coin slides upward on a surface that is inclined at an angle of 16° above the horizontal. The coefficient of kinetic friction between the coin and the surface is 0.23; the coefficient of static friction is 0.31. Find the magnitude and direction of the force of friction under the following circumstances.
(b) after it comes to rest up the incline
.012 N
I've been using my book and trying to get this for about an hour and I'm at a complete loss.
I will be happy to critique your work.
ok well i thought it would be the same as finding the kinetic friction.. since now we're looking for static friction.
first i looked for the normal force. since mg = the coins weight and because we're on an incline i used mgSin(theta) to find the perpendicular force which is the normal force in this case. Then i multiplied it times the coefficient of static friction and i got .012. but evidently that's not right. Except my book doesnt give you a clear description of how to do static friction. kinetic friction yes, but not static.
all correct, except for mgSinTheta. You should be using CosineTheta.
Well, I'm glad you made an effort, but it seems like you got a little mixed up with the trigonometry. Let's break it down step by step.
To find the normal force, we need to use the cosine of the angle since it is the component of the weight perpendicular to the incline. So, the normal force (N) would be equal to mgcos(θ).
Next, we need to find the force of static friction. The equation for static friction is given by Fs ≤ μsN, where μs is the coefficient of static friction.
Given that the coefficient of static friction is 0.31, we can calculate the force of static friction using Fs = μsN.
So, Fs = 0.31 * (mgcos(θ)). Plugging in the values, we have Fs = 0.31 * (0.014 kg * 9.8 m/s^2 * cos(16°)) = 0.055 N.
Therefore, the magnitude of the force of static friction is 0.055 N.
To find the normal force, you are correct that you need to use the weight of the coin, which is given by mg. However, on an inclined plane, the normal force is not equal to the weight. It is the component of the weight perpendicular to the plane, which is given by mg * cos(theta), where theta is the angle of the incline.
So, to find the magnitude of the force of static friction, you would multiply the normal force (mg * cos(theta)) by the coefficient of static friction (0.31), resulting in:
Static friction force = (mg * cos(theta)) * coefficient of static friction
= (0.014 kg * 9.8 m/s^2) * cos(16°) * 0.31
= 0.012 N
Therefore, your answer of 0.012 N is correct. This represents the magnitude of the force of static friction when the coin comes to rest on the incline, and the direction of the force of static friction is up the incline.
To find the magnitude of the force of friction when the coin is at rest and sliding up the incline, you correctly start by finding the normal force. The normal force is the perpendicular force exerted by the surface on the coin.
Since the coin is on an inclined surface, you need to use the component of the gravitational force that is perpendicular to the surface. In this case, you would use the cosine of the angle between the inclined surface and the horizontal. So the normal force is given by:
Normal force = mass of the coin * acceleration due to gravity * cosine(theta)
where mass of the coin is 14 g (convert to kg by dividing by 1000) and acceleration due to gravity is 9.8 m/s^2. Theta is the angle of inclination, which is 16°.
Next, you need to find the force of static friction. The force of static friction is given by:
Force of static friction = coefficient of static friction * normal force
where the coefficient of static friction is 0.31.
So, plug in the values and calculate the magnitude of the force of static friction:
Force of static friction = 0.31 * (mass of the coin * acceleration due to gravity * cosine(theta))
Finally, convert the mass of the coin to kilograms, calculate the product on the right side, and multiply by the coefficient of static friction to get the magnitude of the force of static friction.