Acceleration of a car pulled by a block falling (pulley)

In summary, the car and pulley system setup shown has a car with a mass of 0.05 kg and a block with a mass of 0.02 kg. The car wheels have a friction coefficient of 0.15. Using the free body diagrams and the equations Fnet = ma, Ff = μFn, and Fg = ma, the acceleration of the car can be calculated to be 1.75 m/s^2 in the downward direction. This is in contrast to the previous question where a frictionless surface was assumed and an acceleration of 2.8 m/s^2 was calculated. It is important to note that the direction of positive (+) must be taken into account and in this case,
  • #1
testme
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Homework Statement


A car and pulley system are setup as shown. Find the acceleration of the car. The car wheels have a friction coefficient of 0.15. The car has a mass of 0.05 kg while the block has a mass of 0.02 kg.

http://sadpanda.us/images/836771-GBD7SB9.jpg
*sorry if the picture is bad I tried to sketch it quickly on paint.

I chose up and right to be positive.

Homework Equations


Fnet = ma
Ff = μFn
Fg = ma 3. The Attempt at a Solution
If we draw two separate free body diagrams we would get the following forces.

Car: Force gravity is 0.49 N
Fg = ma = 0.05 * 9.8 = 0.49 N

and since the car isn't accelerating vertically the normal force is the same but opposite direction.

Ff = μFn = 0.49 * 0.15 = 0.0735 N

We also have a tension in the rope that we don't know.

Block: Force gravity is 0.196 N.
Fg = ma = 0.02 * 9.8 = 0.196 N

We also have the tension of the rope which we again don't know.

To calculate the acceleration of the car:

Fnet = ma
Fgcar + Fncar + Ff + T1car + T1block + Fgblock = ma
Since forces cancel out.
Ff + Fgblock = ma
-0.0735 - 0.196 = (0.02 + 0.05) * a
-0.2695 = 0.07 * a
a = -3.85
a = 3.85 m/s^2 [Down]

I have a feeling something is wrong because in the previous question it said there was a frictionless surface and I ended up getting 2.8 m/s^2 for the acceleration which is lower. I'd assume a frictionless surfaces would cause there to be a higher acceleration.

If anyone can verify this for me and tell me what I'm doing wrong it'd be much appreciated.
 
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  • #2
The difficulty is that you are adding the friction force and the force from the block. They are in opposite directions. The force on the car from the falling block is directed to the right while the friction force is directed to the left. Assuming you take "+" to the right, the net force on the car is Fgblock- Ff.
 
  • #3
HallsofIvy said:
The difficulty is that you are adding the friction force and the force from the block. They are in opposite directions. The force on the car from the falling block is directed to the right while the friction force is directed to the left. Assuming you take "+" to the right, the net force on the car is Fgblock- Ff.
So then you mean it'd be like:

Fgblock - Ff = ma

0.196 - 0.0735 = 0.07a
0.1225 = 0.07a
a = 1.75

Also, would I have to follow the curve of the pulley to the right (or clockwise) is positive instead of straight directions, in questions like this?
 

FAQ: Acceleration of a car pulled by a block falling (pulley)

What is the basic concept behind the acceleration of a car pulled by a block falling?

The basic concept behind the acceleration of a car pulled by a block falling is Newton's Second Law of Motion, which states that the force applied on an object is directly proportional to its mass and acceleration. In this case, the block falling exerts a force on the car, causing it to accelerate.

How does the use of a pulley affect the acceleration of the car?

The use of a pulley does not affect the acceleration of the car. The force exerted by the block falling is still the same, regardless of whether it is connected to the car directly or through a pulley. The only difference is the direction of the force, which is redirected by the pulley.

What is the role of friction in this scenario?

Friction plays a minor role in this scenario. It is present between the car and the ground, and between the pulley and the rope. However, these frictional forces are relatively small and do not significantly affect the acceleration of the car.

Can the acceleration of the car be increased by using a heavier block?

Yes, the acceleration of the car can be increased by using a heavier block. This is because according to Newton's Second Law, the force applied on the car is directly proportional to the mass of the block. Therefore, a heavier block exerts a greater force on the car, causing it to accelerate faster.

How can the acceleration of the car be calculated in this scenario?

The acceleration of the car can be calculated using the formula a = F/m, where a is the acceleration, F is the force exerted by the falling block, and m is the mass of the car. The force can be calculated by multiplying the mass of the block by the acceleration due to gravity, and the mass of the car can be measured. Plugging these values into the formula will give the acceleration of the car.

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