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Friction
Questions 2 and 3 refer to a block of mass m pulled with constant velocity over a floor by
a force T inclined at an angle  with respect to the floor as shown below. The coefficient of
friction between the block and the floor is .
2. The magnitude of the vertical component of the force exerted on the block by the floor
is:
a. mg.
b. mg – Tcos
c. mg + Tcos 
d. mg – Tsin 
e. Tsin 


D.
The force exerted by the floor (the normal) is equal to the weight MINUS the amount that
the person is supporting the block. Proof:
F  0
Tsin   N - mg  0
(block is not accelerati ng verticall y)
(N is normal force)
N  mg  T sin 

The magnitude of the frictional force is:



a. Tcos
b. Tsin 
c. 0
d. mg
e. Tcos 
* A. Since the block is neither speeding up nor slowing down, the two horizontal forces
must have the same magnitude, so friction = horizontal component of tension.
From Princeton Review Book
1. A 20 Newton block is being pushed across a horizontal table by an 18 Newton force. If
the coefficient of kinetic friction between the block and the table is 0.4, find the
acceleration of the block. (Use g = 10 m/s2)
a. 0.5 m/s2.
b. 1 m/s2.
c. 5 m/s2.
d. 7.5 m/s2.
e. 9 m/s2.
* C. Watch the difference between mass and weight. The block has a weight of 20 Newtons
and a mass of 2 kg. The 18 Newton force is wining and the friction (8 Newtons) is losing.
Dividing by 2 kg of mass yields 5 m/s2.
3. A crate of mass 100 kg is at rest on a horizontal floor. The coefficient of static friction
between the crate and the floor is 0.4, and the coefficient of kinetic friction is 0.3. A force
F of magnitude 344 N is then applied to the crate, parallel to the floor. Which of the
following is true?
a. The crate will accelerate across the floor at 0.5 m/s2.
b. The crate will slide across the floor at a constant speed of 0.5 m/s.
c. The crate will not move.
d. None of the above.
* C. We need to determine which force is winning: friction or the external 344 N. Calculate
friction: f =  mg = (.4) (100 kg) (10 m/s2) = 400 N. Static friction beats the external force,
but remember this is the upper limit of the cohesive bonds. What this means is that the 344
N is not strong enough to break the cohesive bonds, and the box remains at rest.
13. A block of mass 2 kg slides along a horizontal tabletop. A horizontal applied force of
12 Newtons and a vertical applied force of 15 Newtons act on the block, as shown above.
If the coefficient of kinetic friction between the block and the table is 0.2, the frictional
force exerted on the block is most nearly:
a. 1 N.
b. 3 N.
c. 4 N.
d. 5 N.
e. 7 N.
*E.
The 12 Newton force is a red herring—it has nothing to do with friction except for the fact
that since it makes the block move, we are talking about kinetic friction. Use the Fun Fun
formula:
f  N  (.2)( mg  15 N )  (.2)( 20 N  15 N )  7 N
Note : the normal force must balance the weight plus the additional 15 N of downard force.
16.
(2004, 59%)
A horizontal force F pushes a block of mass m against a vertical wall. The coefficient of
friction between the block and the wall is . What value of F is necessary to keep the block
from slipping down the wall?
a. mg
b.  mg
c.
mg

d. mg (1 - )
e. mg (1 + )
* C.