Physics Question for College Course?

Question

Hey, I need help with 2 problems. The teacher does not teach very well and basically I am trying to learn off the book which is very hard to comprehend. Thanks for your help.

1. An object is hung from a spring balance attached to the ceiling of an elevator. The balance reads F1 = 12N when the elevator is accelerating upward, and reads F2 = 8N when it is accelerating downward with acceleration of the same magnitude a. Find the mass of this object and magnitude of acceleration a. We assume that gravitational acceleration g i known.

2. Darlene the daring, a famous western stuntwoman, is being dragged along the ground by a rope tied to a horse. If her coefficient of kinetic friction with the ground is 1.3 and her mass is 60kg, how much force must the horse pull with to keep her at a constant speed?

Answer 1

(1) The forces on the mass are from the extended spring ( up ) and from gravity ( down ). The net effect is acceleration:

F = ma = kx - mg

F_up = ma = 12 N - mg
F_down = m(-a) = 8 N - mg

Add these equations to get

0 = 20 N - 2 mg
20 N = 2 mg
10 N = mg

so the mass is equal to 10 N / g. Plug that result back into either starting equation to get the acceleration a.

(2) The usual simple model of kinetic friction is F = μ m g where μ is the coefficient of friction. To move at constant velocity, the horse must pull with that same magnitude of force:

F = μ m g

Plug in μ, m, and g to find the horse force.

Answer 2

1) Since the magnitude of the elevator's acceleration is equal in both directions, you know that the force that is acting upward when the elevator goes down and vice versa is equal in magnitude as well (mass doesn't change either so by F=ma, the force must be equal). This means that the force is subtracted from the weight when moving down (you feel light in an elevator when it goes down) and added when you go up (you feel heavier when going up). This means that the weight must be the midpoint between these two scale readings, giveing a force of 2 N and a weight of 10 N. Using F=ma and a = 9.81 m/s^2, you get a mass of 1.02 kg.

For the accleration, you know that the force caused by this accleration is 2 N and that the mass of the object if 1.02 kg, so using F=ma, you can determine that the acceleration is 1.96 m/s^2.

2) Friction problems can get a bit confusing if you try to just look at the equations and not at what is actually going on. As you know, friction is equal to the frictional coefficient times the natural (sometimes called normal) force ( F = (mu)*(F(sub)n) ). The natural force is simply whatever force is pushing the object into the surface (is not acting perpendicular to the surface, you have to find the component of the force that acts perpendicular). In this problem, it's simply the weight of Darlene. What can get confusing is that the frictional force is not always equal to this, but in the case of kinetic friction (friction while moving) the friction is always equal to this. If you draw a Free Body Diagram of the problem, the horse supplies a force in the direction of movement, and the frictional force acts in the opposite direction. For the object to accelerate (move from a standstill), the horse must pull with a greater force than the static (stationary) friction force can supply. To obtain a constant speed, the horse must pull with a force equal to whatever the kinetic friction is. Using the numbers supplied, the kinetic friction is equal to:

F = 1.3(60*9.81)
F = 750 N

Answer 3

1) Inside a 'reference frame' that is accelerating (ie inside the elevator) everything experiences a force indistinguishable from gravity due to that acceleration. This force acts in the opposite direction to the acceleration - this is one of Newton's laws of motion: for every action there is an equal and opposite reaction.

So if the elevator accelerates at a rate of 0.1g (say) everything feels this as a force equal to 10% of the Earth's gravity. Depending on whether the elevator is accelerating up or down, this force will either add or subtract to the planet's gravity - making things heavier or lighter.

We can relate acceleration to force through the equation

F = ma

So if the spring balance shows 12N in one direction and 8N in the other, we can quickly say that at rest the object would show 10N, and the elevator acceleration is +2N one way, and -2N the other, right?

Acceleration due to the Earth's gravity - called g - has a value of about 9.8 m per sec per sec. So we can find the mass of the object

m = F/ a = F / g = F / 9.8 = 10 / 9.8 Kg

Since 9.8 is so close to 10, perhaps the question expects the answer "1 kg". But if you want to calculate m more accurately, go find a calculator.

So the mass is 1kg. When the elevator moves the force it generates on the mass is 2N. So its acceleration is:

a = F / m = 2 N/ 1kg = 2 m per sec per sec

So the acceleration is 2 m per sec per sec, which can be written as 2 ms^-2

2) At a constant speed, with zero acceleration, the force applied by the horse exactly matches the force of friction. Since the coefficient of friction is greater than 1 we know that the sideways force is going to be greater than downwards force of gravity. (a 1.0 coefficient would mean they were the same).

What is the force downwards due to gravity?

F = ma = mg = 60kg x 9.8 ms-2
= 588 N

The coefficient is the ratio of the sideways force to the downwards force, so the horse must pull:

Pull = 588 x 1.3 N
= 764.4 N

Answer 4

I took Intro Physics with Algebra/Trig I and it became an ok type. Alot of the priority contraptions are unique yet once you could observe the uncomplicated suggestions and understand them then you definately are large. i ended up with a intense 80 4%. not undesirable for somebody who began at remedial algebra and not in any respect took trig. I believe the 1st poster that physics is fairly logical. that is not appropriate how good/undesirable you're math. in case you are not getting it, no math theory will ever assist you besides could the "calculus" based intro physics direction could properly be mathematically rigorous yet that would not advise that calculus is used ninety% of the time. From what I heard its basically used for derivations including acceleration and speed, angular momentum and torque, and so on.

Answer 5

m*(g+a)=12N , m*(g-a)=8N , (g+a)-(g-a) = 2a= 4N/m

so a= 2N/m

(g+a)+(g-a) = 2g = 20N/m so g=10N/m and m= 10/g
so a = 2*g/10 = g/5

assuming g=9.81m/s then m= 10/9.81 kg ~1.019 kg.
and a=2/1.019 about 1.962 m/s/s


2. 1.3*60*g Newtons