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50
Dynamics Test 2012
Remember to show your complete work with units, to receive full marks, No work required for Logic Questions.
1. {Logic Question] Consider the following two ways of moving a box across a horizontal floor with constant velocity. In
Case 1, you are pushing on the box while in the Case 2 you are pulling on the box (see pictures). [1 Mark]
(a) The frictional force in case 1 is larger than in case 2.
2. You apply a horizontal force of 500 N to a 100 kg box on a horizontal floor, and you observe that the box has an
acceleration of a = 4 m/s2. [3 Marks, remember to show your work for Part Marks]
What is the kinetic coefficient of friction μK between the box and the floor?
(a) μK = 0.10
F
x
 ma x
F A  F f  max
FA  Ff  ma
500 N  k FN  ma
500 N  uk mg  ma
uk 
500 N  ma
mg
 m
500 N  100kg   4 2 
 s 

m

100kg   9.8 2 
s 

 0.10
The following two questions concern the same physical situation.
3. Picture given below shows the motion of two boxes under the effect of applied force. Friction constant between the
surfaces is uk=0,4. Determine the acceleration of the boxes.
□
F=30N
(a) a= 3.9 m/s
1 kg
 F  ma
3 kg
c
FA  F f  mac

m
m




30 N cos  37   0.4 1kg   9.8 2    0.4    3kg   9.8 2   30 N sin  37     (4kg )a
s 
s 





15.5N   4kg  a
a  3.875
m
s2
4. Determine the tension between the boxes
□
(d) T = 7.8 N
 F  ma
c
FT  F f  mac
FT  mac  F f
m
m


 1kg   3.875 2    0.4 1kg   9.8 2 
s 
s 


 7.795N
5. If the mass of the sleight is 20 kg, and the coefficient of kinetic friction of the snow is 0.23, how much force must the
child apply to move the sleight up the hill (   30 ) with a constant velocity [3 Marks]
□
(a) T = 137 N
 F  ma
c
FA  FG  F f  0
FA  FG  F f
 mg sin  30    mg cos  30 


N
N
  20kg   9.8  sin  30    0.23 20kg   9.8  cos  30 
kg 
kg 


 137 N
This and the following question concern the same physical situation.
In the diagram, the 8kg mass moves up the incline where the coefficient of kinetic friction is 0.4.
8kg
20kg
6. Determine the friction Force acting on the 8 kg mass
F
y
(b) 25.0 N
0
FN  mg cos  37   0
m

FN   8kg   9.8 2  cos  37 
s 

 62.6 N
Ff   FN
 0.4  62.6 N 
 25.0 N
7. Determine the acceleration of each mass
(a) 4.4 m/s2
8kg Block
F
x
20kg Block
 max
FT  FG  Ff  max
m

FT   8kg   9.8 2  sin  37    25.0 N    8kg  ax
s 

FT  72.18 N   8kg  ax
F
y
 ma y
FG  FT  ma y
 20kg   9.8

m
  FT   20kg  a
s2 
196 N  FT   20kg  a
Adding to solve for acceleration
123.82 N   28kg  a
a  4.42
m
s2
8. [Logic Question] A car is moving to the right. A pendulum is suspended from the ceiling and hangs as shown
in the figure. What can we say about the speed of the car?
□ (c) The car’s speed is decreasing
Since there is a (fictitious) force moving the pendulum to the right, the car must be decelerating, thus the speed
is decreasing
9. How much force is needed to give the blocks an acceleration of 5.0 m/s 2 if the coefficient of kinetic
friction between the blocks and the floor is 0.20 ?
4.0 kg
(b) 63 N
3.0 kg
2.0 kg
 F  ma
FA  Ff  ma
FA  ma   FN
 N

N
  9kg   5    0.2  9kg   9.8 
kg 
 kg 

 62.6 N
Question 10
The following three questions concern the same physical situation.
A 30kg traffic light hangs from wires as show in the picture.
a.
Determine the tension T3 [2 Marks]
FT  mg
m

  30kg   9.8 2 
s 

 294 N
b.
Determine the tension T1 [5 Marks]
T1 cos  37  T2 cos  53
T1 sin  37  T2 sin  53   294 N
T1 
T2 cos  53 
cos  37
T2 cos  53
sin  37  T2 sin  53  294 N
cos  37
T2  234.8 N
T1 
234.8 N cos  53 
cos  37 
 176.9 N
Question 11
A window washer pushes his scrub brush up a vertical window at constant speed by applying a force F as shown in
the picture. The brush weighs 12.0 N and the coefficient of kinetic friction is k  0.150 . Calculate:
a) the magnitude of the force F
b) the normal force exerted by the window on the brush.
a)
F
y
 FAy  FG  Ff
0  F sin  53.1   12.0 N  uk F cos  53.1 
F  sin  53.1   0.51cos  53.1    12.0 N
F  16.9 N
b) FN  F cos  53.1 
 16.9 N cos  53.1 
 10.2 N