# Forces & Friction: Acceleration & Coefficient of Friction

• joeG215
In summary, a 1 kg box begins at rest and comes to a stop after a hand applies a force of 2N to it. The coefficient of friction is not given, but the box experiences an accel from rest through 1 meter. The net force on the box is -2N and the coefficient of friction is about .4.
joeG215

## Homework Statement

A 1 kg box begins at rest and a hand applies a constant force of 2N along a 2 meter board. The first meter is frictionless, but the second meter has friction that makes the box come to a stop. Time is not give. Coefficient of friction is not given. My question is, is it possible to find acceleration at the 2nd meter and/or the coefficient of friction?

$F=ma F(f)=uF(n) u= Coef. of friction ## The Attempt at a Solution$F= 2N-F(f) =(1kg)(a)

2N-u(9.8N)=1kg(a) I get stuck with two variables.

Ok so for the first meter you have a box, with a force of 2N being applied to it.

F = ma, so the accel on the box is F/m = 2 m/s^2

Now this box undergoes this accel from rest through 1 meter.

v12=v02+2s(x1-x0)

v12= 0 + 2(2 m / s2)(1m)
v1 = 2m/s

Now, for the box to decelerate from 2 m/s to zero through 1m, you need to have an acceleration of -2 m/s2. You can get this from just doing another kinematic equation or by reason (if it takes 2 m/s^2 to accel from rest to 2m/s through 1 meter, then it should take the same, but opposite to decel to go from 2m/s to rest through 1 meter)

Ok, so now you know that the box must undergo a decel or 2m/s^2 to come to rest after 1 m.

This means that when you sum the forces on the box $$\sum F$$x = max = 1kg* (-2m/s^2)

The net sum must therefore be -2N

The forces acting on the box are the Fhand and the ffriction so sum Fx = Fhand - ffriction = -2N

so ffriction = 4N = $$\mu$$k*N = $$\mu$$k*mg

so $$\mu$$k = about .4

of course.. Thanks

## 1. What is meant by "forces" in the context of physics?

In physics, forces are interactions between two objects that result in a change in motion or acceleration. Forces can be categorized as either contact forces (such as pushing or pulling) or non-contact forces (such as gravity).

## 2. How is acceleration related to forces and friction?

Acceleration is directly related to the net force acting on an object. In the presence of friction, the force of friction acts in the opposite direction of the object's motion, reducing the net force and therefore the acceleration. The amount of friction present also affects the acceleration of an object.

## 3. What is the coefficient of friction and how is it calculated?

The coefficient of friction is a measure of the amount of resistance to motion between two surfaces in contact. It is calculated by dividing the force of friction by the normal force, often represented by the symbol μ. The coefficient of friction can vary depending on the materials and surface conditions of the objects in contact.

## 4. How does the coefficient of friction affect the motion of an object?

The coefficient of friction affects the motion of an object by determining the amount of resistance to motion due to friction. A higher coefficient of friction means there is a greater force opposing the object's motion, resulting in a decrease in acceleration. A lower coefficient of friction means there is less resistance to motion, allowing for a greater acceleration.

## 5. Can friction ever be beneficial in terms of motion?

Yes, friction can be beneficial in certain situations. For example, friction is necessary for walking or driving a car, where the force of friction provides the necessary grip for movement. In some sports, such as ice skating or skiing, friction is intentionally reduced to allow for smoother and faster motion. Additionally, friction can be used to slow down or stop objects in a controlled manner, such as in braking systems.

• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
42
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
416
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
48
Views
6K
• Introductory Physics Homework Help
Replies
16
Views
4K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
16
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K