1.1. These coefficients are just ratios, of force divided by force. So units cancel out.

1.2. Calculate the magnitude of the static friction f_{s}.

Since the static friction (the force that prevents it from moving) is 9N, you need a force greater than 9N to overcome the friction.

Since the static friction (the force that prevents it from moving) is 9N, you need a force greater than 9N to overcome the friction.

1.3. You need to calculate the kinetic friction now. The effect of the static friction is now over.

You need to calculate the mass of the box since this was not given. Only the weight was given.

You need to calculate the mass of the box since this was not given. Only the weight was given.

1.4. Apply a force of 6N.

Since the kinetic friction is 6N, and the applied force is 6N, the net force would be zero. According to Newton's First Law, the box would move at constant velocity due to it's inertia.

Since the kinetic friction is 6N, and the applied force is 6N, the net force would be zero. According to Newton's First Law, the box would move at constant velocity due to it's inertia.

Draw a graph of

1.5.

- As the applied force is increased, the static friction force increases too, up to a maximum of 9N.

Static forces increase as applied force is increased.

As the applied force is increased from 0 to 9N, the static friction increases. - After the applied force overcomes the static force (after 9N), a new friction force (kinetic friction) becomes relevant.
- Unlike static friction, kinetic friction remains constant as applied force is increased.

The applied force is being increased from 9N to 20N, but the kinetic friction remains constant at 6N.

2.

Since the vehicle stops in 34,01m, the car does not hit the object which was 35m away.

- Convert 72 km.h
^{-1}to m.s^{-1}by dividing by 3,6. - Find weight and normal force
- Calculate kinetic friction force.
- Calculate acceleration from F
_{net}= ma - Calculate actual stopping distance
- Compare to available distance

Since the vehicle stops in 34,01m, the car does not hit the object which was 35m away.

3.1.

3.2. Calculate the maximum static friction.

The blocks weight is only 5,7N and is unable to overcome this maximum static friction of 7,8N. Hence the block does not move.

The blocks weight is only 5,7N and is unable to overcome this maximum static friction of 7,8N. Hence the block does not move.

3.3.

5,7N

Static friction can have many values, up to it's maximum. It's maximum in this example is 7,8N. But it is only exerting enough friction force to balance the weight of 5,7N.

5,7N

Static friction can have many values, up to it's maximum. It's maximum in this example is 7,8N. But it is only exerting enough friction force to balance the weight of 5,7N.

What is the minimum applied force, that would first allow the block to just start sliding downwards?

3.4. The weight is 5,7N. So the friction must become LESS than 5,7N. Therefore calculate the normal force that would produce a friction force equal to the weight of 5,7N. Then use a force slightly less to get a friction slightly less.

This means that any applied force less than 9,5N would produce a static friction force less than 5,7N. And the block would not be able to be supported, and would slide down.

This means that any applied force less than 9,5N would produce a static friction force less than 5,7N. And the block would not be able to be supported, and would slide down.

3.5.

Since the system is now moving, remember to use the coefficient of kinetic friction to calculate friction.

For constant velocity, the vertical forces must be balanced. Hence the friction force must equal to the weight.

Apply a force of 11,4N

If you had applied a force of 11,4N at the beginning, before the block started moving, the system would have remained at rest.

You would use the static coefficient of 0,6 and get a friction of 6,84N which would be larger than the weight. And the block would not be moving!

So you need a little extra force to overcome static friction! Then it is easier to move.

Since the system is now moving, remember to use the coefficient of kinetic friction to calculate friction.

For constant velocity, the vertical forces must be balanced. Hence the friction force must equal to the weight.

Apply a force of 11,4N

If you had applied a force of 11,4N at the beginning, before the block started moving, the system would have remained at rest.

You would use the static coefficient of 0,6 and get a friction of 6,84N which would be larger than the weight. And the block would not be moving!

So you need a little extra force to overcome static friction! Then it is easier to move.