Newton, freefall acceleration problem

• bengt2665
In summary, the weight of the cart and hanging mass is 4 N each. The friction of the cart on the track is a constant force of 2 N. The mass of the string, pulley, and friction of the pulley can be ignored. To find the magnitude of acceleration of the cart in terms of free-fall acceleration, we use the equations T - Ff = ma and mg - T = ma and substitute the given masses of 4 N for both the cart and hanging mass. Solving for T and substituting in the equations, we get a = g/4 as the final answer.
bengt2665
The weight of the cart is 4 N. The weight of the hanging mass is 4 N.
Friction of the cart on the track is a constant force of 2 N magnitude.
The mass of the string, the mass of the pulley, and the friction of the pulley can be neglected.
What is the magnitude of the acceleration of the cart if g is the free-fall acceleration?
Picture attached.

The answer is to be given in terms of g.

Okay, here's what I have so far:

By drawing a free body diagram I manage to get these two equations for the forces acting on both the cart and the hanging mass:

I'm assuming positive acceleration to the right of the cart and down for the mass.

For the cart: T - Ff = ma
For the mass: T - mg = ma
(since they have the same weight, there is only one m)
We know that the weight of these two is 4N each so for both, 4 = ma (i'm predicting this might not be right...)

I'm having trouble applying their given information (4N weight for both masses) with the equations I'm coming up with, for example I don't know what this 4N corresponds to...
Is it 4N = m.a for the cart and 4N = mg for the hanging mass?

How do I apply the information they are giving me?

Thanks!

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For mass ma = mg - T.
For cart 4N is the normal reaction. In this problem it is not needed.
For the cart: T - Ff = ma
For the mass: mg- T = ma .Solve for T.
Hence find a by substituting m = 4/g

This doesn't really help me understand...
I end up having 2 equations:
1) T - Ff = ma
2) -T + mg = ma

Then I end up with

mg - Ff = 2ma (by combining 1 and 2)

and I'm stuck there.

Just figured it out...

we know that m1g=4 and m2g=4 => m1 = 4/g, m2 = 4/g

So in the end, after substituting in 1 and 2 we get

1) T1 - 2 = (4/g)a
2) -T1 + 4 = (4/g)a

By adding these 2 equations (to get rid of the T) we end up with
2 = 8a/g
which gives a = g/4.

1. What is the freefall acceleration problem?

The freefall acceleration problem is a physics problem in which an object is dropped from a certain height and its acceleration due to gravity is measured. The goal is to calculate the freefall acceleration, which is a constant value for objects in a vacuum, using Newton's second law of motion.

2. Who was Sir Isaac Newton and why is he associated with the freefall acceleration problem?

Sir Isaac Newton was a famous English scientist and mathematician who is known for his contributions to the fields of physics and mathematics. He is associated with the freefall acceleration problem because he developed the laws of motion, particularly the second law which is used to solve this problem.

3. How is the freefall acceleration problem solved?

The freefall acceleration problem is solved by using the formula a = F/m, where a is the acceleration, F is the net force acting on the object, and m is the mass of the object. The net force is calculated by subtracting the weight of the object from the force of gravity, and the mass is typically given in the problem.

The units of measurement for freefall acceleration are meters per second squared (m/s^2). This represents the change in velocity (m/s) over the change in time (s).

5. How is the freefall acceleration problem related to the law of universal gravitation?

The freefall acceleration problem is related to the law of universal gravitation because both involve the force of gravity. In the freefall acceleration problem, the force of gravity is used to calculate the acceleration of an object, while the law of universal gravitation describes the force between two objects due to their masses and the distance between them.

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