A student sets up a experiment similar to one we did in lab. A 0.38 kg wood bloc

Question

A student sets up a experiment similar to one we did in lab. A 0.38 kg wood block is placed on a horizontal lab track and has a 1.5 kg brass mass placed on top of it. The wooden block is attached to a force probe via a light string similar to what we did in lab. The block is initially at rest and she slowly increases the tension in the string until the mass starts to move. She then pulls the block along at a constant velocity of 1.5 m/sec. The signal from the force probe produces the graph below on the computer. a) What is the co-efficent of static friction between the wooden block and the lab track? mu s = 2.82 NOb)What is the co-efficent of kinetic friction between the wooden block and the lab track? mu k = 2.82 NOc) What would the force probe read if the student pulled the block along at a constant speed of 3 m/sec instead of 1.5 m/sec as she initially did ? F= N d) What would the force probe read if the student placed a 3 kg brass mass on the block instead of the 1.5 kg mass she initially used ? F= N

Explanation / Answer

1) To find coefficient of static friction, take the maximum static frictional force and divide by the normal force.
Ff =Fn

1.5+.388 = 1.88 *9.8 = 18.42N

7N/18.42N = 0.38

2. To find the coefficient of kinetic friction, take the average force due to kinetic friction and divide by the normal force

6N/18.42N = 0.326

3.If she is pulling it at a constant speed, there is no acceleration so the frictional force must still equal the applied force so the force probe will still read 6N

4. I am assuming there is no acceleration and that she is pulling with constant speed again. If there is acceleration that would change things. The new normal force = 3 + .388 = 3.388kg * 9.8 = 33.2N

Ff = 0.326 * 33.2N = 10.82N - kinetic frictional force.

Ff = 0.38 * 33.2N = 12.62N - maximum static frictional force

Hope that helps

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