Calculate Acceleration of 5kg Box w/ μ=.5 & 2kg Mass Falling

  • Thread starter Ozball
  • Start date
In summary, when calculating the acceleration of a 5kg box attached to a 2kg mass by a massless cord over a frictionless pulley with a coefficient of μ=.5, you can use the formula a = (Fnet/total mass) where Fnet is the net force acting on the system. The coefficient of μ represents the amount of friction present, with a smaller μ resulting in easier movement and a larger μ resulting in more difficult movement. There are two types of μ values for an object: static, which is the resistance to acceleration, and kinetic, which is the resistance to acceleration after the object starts moving.
  • #1
Ozball
1
0
1. Calculate the acceleration of a box with a mass of 5kg and μ=.5 when attached by a massless cord over frictionless pulley to a 2kg mass that is allowed to fall freely. The box is on a table, the pulley is hanging off.



2. I know a = (Fnet/total mass) but I am not sure what to do with μ! Can someone explain it to me?
 
Physics news on Phys.org
  • #2
When you pull a sled on ice, it is easy to pull because μ is small.
When you pull it on gravel, it is hard to pull because μ is larger.
μ is the fraction of the weight that the friction force is.
Force of friction = μ*weight.
(When on a slope at an angle, you have to use the normal force - the component of the weight that pushes against the slope.)
 
  • #3
Think of mu (μ) as the magnitude of how much an object cancels out the forces trying to accelerate it.

There are two mu values for any object. The static and kinetic friction coefficients. Static is the resistance for an object to acceleration, kinetic is the resistance to acceleration after it starts moving.
 

FAQ: Calculate Acceleration of 5kg Box w/ μ=.5 & 2kg Mass Falling

How do I calculate the acceleration of a 5kg box with a coefficient of friction of 0.5 and a 2kg mass falling?

To calculate the acceleration of an object, we use the formula a = (m1g - μm2g)/m1, where m1 is the mass of the object, μ is the coefficient of friction, m2 is the mass of the falling object, and g is the acceleration due to gravity (9.8 m/s2). Plugging in the values, we get a = (5kg x 9.8 m/s2 - 0.5 x 2kg x 9.8 m/s2)/5kg = 4.51 m/s2.

How does the coefficient of friction affect the acceleration of the box?

The coefficient of friction is a measure of the resistance to motion between two surfaces. The higher the coefficient of friction, the greater the force needed to move the box. In this case, a higher coefficient of friction means a lower acceleration, and vice versa.

What is the significance of the 2kg mass falling in the calculation?

The 2kg mass falling adds an additional force acting on the 5kg box due to gravity. This force is equal to m2g, where m2 is the mass of the falling object and g is the acceleration due to gravity. This additional force affects the overall acceleration of the box.

Can the acceleration of the box be negative?

Yes, the acceleration of the box can be negative if the force of friction is greater than the force of gravity. This would result in a deceleration or negative acceleration.

How does the acceleration of the box change if the coefficient of friction is increased?

If the coefficient of friction is increased, the acceleration of the box will decrease. This is because there is a greater resistance to motion between the surfaces, resulting in a smaller net force and therefore a smaller acceleration.

Similar threads

Acceleration and Tension in a Pulley System
Replies
9
Views
5K
Why body A has greater acceleration?
Replies
13
Views
779
Force problem with three masses and two pulleys
Replies
4
Views
2K
Need help calculating the tension in the string of two boxes with friction
Replies
1
Views
6K
Experimental Design: Pulley and Mass Hangers
Replies
30
Views
2K
Calculating tensions and acceleration
Replies
10
Views
2K
Calculate Tension in the String
Replies
24
Views
9K
Multi-Mass Force/Acceleration problem
Replies
5
Views
3K
Coefficient of Friction of Fixed Atwood Pulley
Replies
1
Views
7K
How can the tension in a wire change on either side of a pulley?
Replies
2
Views
8K

Hot Threads

Recent Insights

Back
Top