Question

6 multiple choice 1 point
a truck driver frequently loses her phone when it slides off of the dash. she wants to find a phone case that wont slide as easily. having studied physics in school she decides to make her purchase based off of the phone cases coefficients of static friction. which coefficient of friction listed below would give her phone the most grip?
0.05
0.1
0.5
1.2
7 formula 1 point
a student pushes on a heavy refrigerator until the fridge is just about to move. at this point the student is pressing against the fridge with a force of 352 newtons. what must be the magnitude of the fridges maximum static friciton force?
answer

8 formula 1 point
a object feels a normal force of 230 newtons and a kinetic friction force of 58 newtons. what must be the coefficient of kinetic friction for the object?
answer

9 formula 1 point
a physicist applies a force against a box. the box is just about to move when the applied force is 44 newtons. if the box has a mass of 4 kg, what must be the coefficient of static friction for the box?
answer

Explanation:

Question 6

Step1: Understand static friction

The maximum static friction force \(F_{s}=\mu_{s}N\), where \(\mu_{s}\) is the coefficient of static friction and \(N\) is the normal force. A larger \(\mu_{s}\) means a larger maximum static friction force (more "grip").

Step2: Compare values

We compare the given values \(0.05\), \(0.1\), \(0.5\), and \(1.2\). Since \(1.2>0.5 > 0.1>0.05\)

Step1: Use equilibrium condition

When the fridge is just about to move, the applied force \(F_{applied}\) is equal to the maximum static friction force \(F_{s,max}\) (from Newton's first law, \(F_{net}=0\) at the verge of motion)

Step1: Recall kinetic - friction formula

The formula for kinetic friction is \(F_{k}=\mu_{k}N\), where \(F_{k}\) is the kinetic - friction force, \(\mu_{k}\) is the coefficient of kinetic friction, and \(N\) is the normal force.

Step2: Solve for \(\mu_{k}\)

We can rewrite the formula as \(\mu_{k}=\frac{F_{k}}{N}\). Given \(F_{k} = 58\space N\) and \(N=230\space N\), then \(\mu_{k}=\frac{58}{230}\approx0.25\)

Answer:

1.2

Question 7