Download Kinematics Multiples

Inclines
From Princeton Review Book
1. A frictionless inclined plane of length 20 m has a maximum vertical height of 5 meters.
If an object of mass 2 kg is placed on the plane, which of the following best approximates
the net force it feels? Use g = 10 m/s2. Hint—Use information given to find sine or cosine
as ratios. No need to use inverse trig functions.
a. 5 N.
b. 10 N.
c. 15 N.
d. 20 N.
e. 30 N.
* A. The NET force is only (mgsin). Mgcos is canceled by the normal force. Don’t make
too much work for yourself. The sine of the incline is OPP/HYP = 5m/20m. Thus, mgsin
= (2 kg) (10 m/s2) (5m/20m)= 5 Newtons.
2. The coefficient of static friction between a box and a ramp is 0.5. The ramp’s incline
angle is 30 degrees. If the box is placed at rest on the ramp, the box will:
a. accelerate down the ramp.
b. accelerate briefly down the ramp but then slow down and stop.
c. move with constant velocity down the ramp.
d. not move.
e. Cannot be determined from the information given.
*A. Draw and FBD!!! The force pulling the box down the ramp is the component of weight
parallel to the incline: mgsin. The force resisting the motion is static friction: mgcos.
So, the question is: which force is greater— mgsin or mgcos ? Calculate each, leaving
m and g as variables: mgsin = mgsin(30) = mg(1/2)=mg/2.
mgcos = (.5) (mg)cos(30) = ( 12 )(mg)( 23 ) .
Clearly, the friction is less than the component of the weight down the incline.
This means the box will begin to slide.
Plus, since kinetic friction tends to be less than static friction, it will continue to slide at a
constant acceleration.
Questions 3 and 4 refer to the following figure:
3. If the surface of the incline is frictionless, how long will it take the block to reach the
bottom if it was released from rest at the top?
a. 0.7 seconds.
b. 1.4 seconds.
c. 2.0 seconds.
d. 2.8 seconds.
e. 4.0 seconds.
*D. This is a really good review of thinking quick on your feet with trig functions. You
know that the sine of 30 degrees is ½, so the hypotenuse of the triangle must be 20 meters.
The acceleration of an object down an incline is g*sin() = (10 m/s2)(1/2) = 5 m/s2.
Then, use position function:
d
1 2
at
2
20 
1
(5)t 2
2
t  2.83 sec onds
4. If the surface of the incline is frictionless, with what speed will the block reach the
bottom if it was released from rest at the top?
a. 8 m/s.
b. 10 m/s.
c. 14 m/s.
d. 18 m/s.
e. 20 m/s.
*C. Since the object fell from rest:
v  vo  at  0  (5m / s 2 )( 2.83 sec)  14m / s
From Old AP’s
Questions 5 and 6
A block slides down a frictionless incline as indicated in the figure below. Let Y = 0, X =
0, and t = 0 just as the block is released from rest at the top of the incline. Let the positive
sense for Y be up and the positive sense for X to be to the right. Note: this is NOT the
rotated axis system we used in class, which makes this problem tricky.
5. Which of the following graphs best represents the X-component of the velocity of the
block as a function of time?
a.
b.
c.
d.
e.
*E. If you look carefully at the diagram, the block is moving in the positive x direction.
Don’t be confused by the fact that the axis does not line up with the incline. The
acceleration can be broken down into a constant X-component and a constant Ycomponent. Thus, we want an acceleration that is increasing uniformly in the positive
direction.
6. Which of the following graphs best represents the X-component of the position of the
block as a function of time?
a.
b.
c.
d.
e.
* C The acceleration is constant and positive, which means the position function is a
parabola. Even though the acceleration has an X and a Y component, each component is
constant.
Questions 7 and 8:
A 2 kg block, starting
from rest,
slides 20 m down a frictionless inclined plane from X to Y, dropping a vertical distance of
10 meters, as shown above.
7. The magnitude of the net force on the block while it is sliding is most nearly:
a. 0.1 N.
b. 0.4 N.
c. 2.5 N.
d. 5.0 N.
e. 10.0 N.
* E. The normal force and mgcos are canceling. There is no friction. Thus, the only
unbalanced force is the component of the weight parallel to the incline:
F  mg sin   (2kg)(10m / s 2 )(10m / 20m)  10 N
Note: since you cannot have a calculator on the multiples, it is a really handy trick to use
the actual ratio of opp./hyp. to calculate the sines and cosines.
8. The speed of the block at point Y is most nearly:
a. 7 m/s.
b. 10 m/s.
c. 14 m/s.
d. 20 m/s.
e. 100 m/s.
*C. This is a combined incline/kinematics problem. First calculate the acceleration of the
block, then use Don’t Know, Don’t Care to find the speed. (Later on, you could also use
conservation of energy).
a  g sin   (10m / s 2 )(10m / 20m)  5m / s 2
V 2  V02  2a( X  X 0 )
V 2  0  2(5m/s 2 )( 20m  0)
(define down the incline as positive)
V  200  14.14m / s
9. An object released from rest at time t = 0 slides down a frictionless incline a distance of
1 meter during the first second. The distance traveled by the object during the time
interval from t = 1 second to t = 2 seconds it:
a. 1 m.
b. 2 m.
c. 3 m.
d. 4 m.
e. 5 m.
* C. This is actually a kinematics problem once you realize that the presence of the incline
is only to tell you that acceleration is constant and not free fall. You must first calculate the
acceleration, then find the distance traveled in two seconds and then subtract the distance
traveled in the first second:
d  12 at 2
1m  12 a(1s ) 2
So accelerati on  2 m/s 2
Now, find the distance traveled from t  0 to t  2 :
d  12 (2m / s 2 )( 2 s ) 2  4meters
So, actual distance in time interval is 4m - 1 m  3 meters.
Note: you have to use the entire 2 seconds in the second equation because if you want to
restart the clock at t = 1 second, you need to find the new initial velocity and add an
additional term—which you can do if you like.
10. (1993) A block of mass 5 kg lies on an inclined plane, as shown below. The horizontal
and vertical supports for the plane have lengths of 4 meters and 3 meters, respectively.
The coefficient of friction between the plane and the block is 0.3. The magnitude of the
force F necessary to pull the block up the plane with constant speed is most nearly:
a. 30 N.
b. 42 N.
c. 49 N.
d. 50 N.
e. 58 N.
* B Draw an FBD!!! In order to be in equilibrium, the force must equal the sum of the
component of weight down the incline plus the force of friction. Also, note that this is a 34-5 triangle and use this fact to set up ratios for the sine and cosine.
Questions 11 – 12:
A plane 5 meters in length is inclined at an angle of 37 degrees as shown above. A block of
weight 20 N is placed at the top of the plane and allowed to slide down.
11. The mass of the block is most nearly:
a. 1.0 kg.
b. 1.2 kg.
c. 1.6 kg.
d. 2.0 kg.
e. 2.5 kg.
*D. Since W=mg and g is about 10 m/s2, m is about 2 kg. The incline has no effect on this.
15. The magnitude of the normal force exerted on the block by the plane is most nearly:
a. 10 N.
b. 12 N.
c. 16 N.
d. 20 N.
e. 33 N.
*C. Use the triangle shown and not your calculator to calculate the cosine:
N  mg cos   (20 N )(
4m
)  16 N
5m
16 through 18.
A small box is on a ramp tilted at an angle  above the horizontal. The box may be subject
to the following forces: Friction (f), gravitational (mg), pulling or pushing (Fp) and normal
(N). In the following free-body diagrams for the box, the lengths of the vectors are
proportional to the magnitudes of the forces.
16.
(2004, 96 %)
Which figure best represents the free-body diagram for the box if it is accelerating up the
ramp?
a. Figure A.
b. Figure B.
c. Figure C.
d. Figure D.
e. Figure E.
* E. If it is accelerating up the ramp, there must be a net force up the ramp. This knocks out
diagram A.
Figure B is incorrect because the normal is vertical and not perpendicular to the ramp.
Figure C is in equilibrium if you look closely at the magnitude of friction vs. mgsin.
Figure D is incorrect because the breakdown of mg is wrong.
Figure E is correct--the net force is up the ramp, the motion is up the ramp, and therefore
the friction is down the ramp.
17.
(2004, 54 %)
Which figure best represent the free-body diagram for the box if it is at rest on the ramp?
a. Figure A.
b. Figure B.
c. Figure C.
d. Figure D.
e. Figure E.
* C. At rest means equilibrium
Friction is balancing mg sin .
Normal is balancing mg cos 
18.
(2004, 44%)
Which figure best represents the free-body diagram for the box if it is sliding down the
ramp at constant speed?
a. Figure A.
b. Figure B.
c. Figure C.
d. Figure D.
e. Figure E.
* C. Trick question. Constant speed means the box is still in equilibrium.
Friction is balancing mg sin .
Normal is balancing mg cos 