Home

FRICTION PROBLEMS

*Remember
F_n = Weight = mg
on horizontal surface

Ex) A 5.0 kg Steel block is resting on a *horizontal table*.  The coefficient of static friction (u_s) is 0.75 and the Uk is 0.57.

a) What minimum force is needed to start this block moving (fs)?

Ex) A 5.0 kg Steel block is resting on a *horizontal table*.  The coefficient of static friction (u_s) is 0.75 and the U_k is 0.57.

a) What minimum force is needed to start this block moving?

f_s= F_nu_s = wu_s = mgu_s

f_s = mgu_s

= 5.0 kg(10. m/s²)(.75)

= 38. N

b) What is the frictional force on this object as it moves?

f_k = mgu_k

= 5.0 kg(10. m/s²)(.57)

= 29 N

c) What force must be applied to the object to keep it moving at constant velocity? (Hint: F_net = 0)

F applied =  friction force

= 29 N

Friction Questions

Coefficients of Friction & F = ma

Remember ...

Static Friction    f_s = U_sF_n

Kinetic Friction  f_k = U_kF_n

Ex 1) A 1000 N car skids on a wet concrete road. If the road is horizontal, what is the friction force on the car?

Ex 1) A 1000. N car skids on a wet concrete road. If the road is horizontal, what is the friction force on the car?

w = 1000. N

U_k  = .58

f_k  =

f_k = U_kW = .58 x 1000. N

= 5.8 x 10² N

Ex 2) A skier is being pulled along a horizontal surface at constant speed with a force of 50. N.

 What is the weight of the skier?

Ex 2) A skier is being pulled along a horizontal surface at constant speed with a force of 50. N.

 What is the weight of the skier?

At constant speed:
f_k = applied force

f_k  = 50. N

w = ?

u_k = .05

f_k =u_kw

50 N = .05W  

w = 1000 N

Ex 3) A 10. kg block of wood sliding on a horizontal wooden table is brought to rest. What is the force on the block of wood that caused it to stop?

Ex 3) A 10. kg block of wood sliding on a horizontal wooden table is brought to rest. What is the force on the block of wood that caused it to stop?
 

The force of kinetic friction
stops the block.

f_k = ?

U_k  = .30

f_k= F_nu_k = wu_k = mgu_k

f_k=mgu_k

f_k = (10. kg)(10 m/s²).30 =30. N

Ex 4) A 53. kg block, slowed by friction, has an acceleration of -0.1 m/s². What is the force of friction on the block?

Ex 4) A 53. kg block, slowed by friction, has an acceleration of -0.1 m/s². What is the force of friction on the block?

m = 53. kg

a = -.10 m/s²

f_k = F = ma

f_k = F = 53 kg(-0.1 m/s²) 

= -5.3 N

Ex 5) A force of 8.0 N gives a 3.0 kg mass an acceleration of 2.0 m/s²  to the right. What is the friction on the block?

F_applied = 8.0 N Right
m = 3.0 kg
a = 2.0 m/s²  Right

F_net = ma = 3.0 kg(2.0 m/s² )

= 6.0 N

What friction force will produce a net force of 6.0 N?