can you please help me with these True False A B C Q3: Is the following statemen

Question

can you please help me with these

True

False

A

B

C

Q3: Is the following statement true or false? The coefficient of static friction can only be used to calculate the maximum static friction force.

True

False

a. Increase

b. Decrease

c. Stay the same

Q5:The following numbers show the kinetic friction force on a mass being dragged on a horizontal surface. Which of the following has the highest kinetic friction coefficient?

Q6: In this experiment, if the recorded friction force is doubled while the contact surface remains the same. Which of the following, if any, has possibly also been doubled?

The frictional coefficient

Total area in contact

Mass of the object

None of the above

True

False

Q6: In this experiment, if the recorded friction force is doubled while the contact surface remains the same. Which of the following, if any, has possibly also been doubled?

a.

The frictional coefficient

b.

Total area in contact

c.

Mass of the object

d.

None of the above

Q7: Is the following statement true or false? Assuming the simplified friction model, , is always correct, contact area should not affect the frictional force.

True

False

Q8: Based on the equation, , which of the following is true if the friction force become doubled for the same surface?

a.

doubles

b.

triples

c.

doubles

d.

triples

Explanation / Answer

Q1) static frictional force is Fs = mu_s*N

So the answer is TRUE

Q2) Kinetic frictional force = mu_k*m*g

So best representation is C

Q3) Yes ,we can calculate maximum static frictional force by using coefficient of static friction

So the answer is TRUE

Q4) static coefficient is a consatnt between two bodies which are in contact

so the answer is C) stay the same

Q5) coefficient of kinetic friction is mu_k = f_k/(m*g) =

for f_k = =10 N and m = 10 kg ,then mu_k = 10/(10*9.8) = 0.102

For f_k = 40 N and m = 20 kg then mu_k = 40/(20*9.8) = 0.204

for f_k = 5N and m = 1 kg then mu_k = 5/(1*9.8) = 0.51

for fk = 10 N and m = 5 kg then mu_k = 10/(5*9.8) = 0.204

So the answer is f_k = 5N and mass m = 1 kg

Q6) None of the above

Q7) contact area should effects the frictional force

So the answer is False

Q8) mu_s is consatnt here in this case

so the answer is c) N doubles

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