Ranking Task
FRICTION PROBLEMS
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_{n}u_{s} = F_{g}u_{s} = mgu_{s}
f_{s} = mgu_{s}
= 5.0 kg(9.8 m/s^{2})(.75)
= 37. N
b) What is the frictional force on this object as it moves?
f_{k} = mgu_{k} = 5.0 kg(9.8 m/s^{2})(.57)
= 28 N
c) What force must be applied to the object to keep it moving at constant velocity? (Hint: F_{net} = 0) At Constant velocity
F applied = friction force
= 28 N
Friction Questions
Coefficients of Friction & F = ma
Remember ... Static Friction f_{s} = U_{s}F_{n} Kinetic Friction f_{k} = U_{k}F_{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?
w = 1000. N U_{k } = .58 f_{k } =
f_{k} = U_{k}F_{g} = .58 x 1000. N
= 5.8 x 10^{2} 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?
At constant speed:
f_{k } = 50. N w = ?
u_{k} = .05 f_{k} =u_{k}F_{g} 50. N = .05[F_{g}] 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? The force of kinetic friction f_{k }= ?
U_{k }= .30
f_{k }= F_{n}u_{k} = F_{g}u_{k} = mgu_{k}
f_{k }= mgu_{k}
f_{k}
= (10. kg)(9.8 m/s^{2}).30 = 29. N
Ex 4) A 53. kg block, slowed by friction, has an acceleration of 0.1 m/s^{2}. What is the force of friction on the block?
m = 53. kg a = .10 m/s^{2}
f_{k} = F = ma
f_{k}=F = 53. kg(.10 m/s^{2})
= 5.3 N
Ex 5)
A force of 8.0 N gives
a
3.0 kg
mass an acceleration of 2.0 m/s^{2}
to the
right. F_{applied}
= 8.0 N Right
m = 3.0 kg
a = 2.0 m/s^{2} Right
F_{net} = ma = 3.0 kg(2.0 m/s^{2} )
= 6.0 N
What friction force will produce a net force of 6.0 N?
friction = 2 N left
Review ***AP Pulleys***
ŠTony Mangiacapre.,  All Rights Reserved

