Download Kinematics Multiples

Conservation of Energy
From Princeton Review Book
1. A box of mass m slides down a frictionless inclined plane of length L and vertical
height h. What is the change in its gravitational potential energy ( Ug)?
a. –mgL
b. –mgh
c. –mgL/h
d. –mgh/L
e. –mghL
* B. Since U = mg(hf = hi) = mg(0 – h) = -mgh. The length of the incline is not
important—only the vertical height through which the block traveled. That is the beauty
of potential energy functions: you only need to know initial and final conditions.
2. A particle is subjected to a conservative force whose potential energy function is :
U ( x)  ( x  2) 3  12 x
where U is in Joules when x is measured in meters. Which of the following represents a
position of stable equilibrium?
a. x = -4.
b. x = -2
c. x = 0.
d. x = 2.
e. x = 4.
* E. Equilibrium occurs when the force is zero.


dU
  3( x  2) 2  12  0
dx
Or : (x - 2) 2  12 / 3  4
F 
So : (x - 2)  4, or x  2  2
This means x  0 or 4 which are both choices.
This suggests that one point is a stable equilibriu m the other unstable.
Stable equilibriu ms occur at minimums of functions which is when the second
derivative is positive (so the function is concave up).
U ' '  6 ( x  2)
U " (0)  12 which means x  0 is a maximum (unstable equilibriu m)
U " (4)  12, which means x  4 is a minimum and stable equilibriu m
3. Person A throws a stone of mass m with initial velocity Vo at an angle  with the
horizontal from a bridge that is H meters above the water level. At the same instant,
Person B throws a stone of mass m with the same initial speed in the horizontal
direction as shown below.
Which of the following is the correct relationship between the speeds of each stone VA
and VB and the relationship between their total time in the air: tA and tB?
a. VA < VB
and tA > tB
b. VA > VB
and
c. VA < VB
and tA = tB
tA < tB
d. VA = VB
and tA > tB
e. VA = VB
and tA = tB
* D. Since both stones have the same KE and the same Ug at the start, they have the same
TME. Energy is a scalar, so it does not matter that they are not headed in the same
direction. Thus, by conserving energy, they must have the same speed when they impact
the water (although they will not have the same impact angle). However, since stone A is
thrown upward, it will take longer to hit the water.
From Old AP’s
4. (1974) For a block of mass m to slide without friction up a rise of height h, the
minimum initial speed of the block must be:
a.
gh
2
b.
gh
2
c.
2 gh
d. 2 gh
e. 2 2 gh
*C. Apply the conservation of energy:
TME A  TME B
KE A  U gA  KE B  U gB
1
2
mV 2  0  0  mgh
(to just make it up the hill, the KE will be zero at top.)
V  2gh
5. (1993) A ball is thrown upward. At a height of 10 meters above the ground, the ball
has a potential energy of 50 Joules (with the potential energy equal to zero at ground
level) and is moving upward with a kinetic energy of 50 Joules. Air friction is
negligible. The maximum height reached by the ball is most nearly:
a. 10 m.
b. 20 m.
c. 30 m.
d. 40 m.
e. 50 m.
*B.
TME  U  KE  50 J  50 J  100 J
The maximum height occurs when all KE has ben converted to U.
Since at 10 meters, half of the energy is potential, at 20 meters all of the energy
will be stored. Note - -this is only true becuase gravitatio nal potenial energy is linear wit h height.
This would not be true for elastic potential energy.
You can also calculate the answer :
At h  10 meters : U  50 Joules
mgh  50 Joules
mg(10m)  50Joules
so mg  5 Newtons.
TME peak  mghmax
100 J  (5 N )h
so h  20 meters
6. A block of mass m slides on a horizontal frictionless table with an initial speed Vo. It
then compresses a spring of force constant K and is brought to rest. How much is the
spring compressed from its natural length?
a.
V02
2g
b.
mg
V0
K
c.
m
V0
K
d.
e.
m
V0
K
K
V0
m
*D. Conserve energy: the energy is initially all KE, and when it is brought to rest has
been converted to elastic potential energy:
TME A  TME B
1
2
mV02  12 K (x) 2
So : x 
mV02

K
m
V0
K
7. (1974) A block attached to a spring of negligible mass is released from rest on a
frictionless surface with an initial elastic potential energy of 50 Joules. When the
displacement of the block is half the maximum value, its instantaneous KE is:
a. zero.
b. 12.5 Joules.
c. 25 Joules.
d. 37.5 Joules.
e. 50 Joules.
*D. This one is a little tricky.
Using logic: The TME is 50 Joules. When the displacement of the block is cut in half, the
elastic potential energy is cut in FOURTH because the displacement is squared in the
formula. This means Us = 12.5 Joules, leaving 32.5 Joules left for KE.
You could also make up a value for the spring constant—make it easy on yourself and
call it 1 N/m.
Then solve the problem using formulas:
U s  12 K (x) 2
50 J  12 (1N / m)( x) 2
so X  10 meters at max. stretch
TME A  TME B
50 Joules  KE  12 (1N / m)(10m / 2) 2
KE  37.5 Joules
8. The potential energy of a body of mass m is given by:
U   mgx 
1
2
Kx 2
The force acting on the body at position x is:
a. 
mgx 2
Kx 3

2
6
mgx 2
Kx 3
b.

2
6
c.  mg  12 Kx
d.  mg  Kx
e.
mg - Kx
* E.
F -
dU
 - - mg  Kx  mg  Kx
dx
9. (1984) If a particle moves in such a way that its position X is described as a function of
time t by:
X t
3
2
then its kinetic energy is proportional to:
a. t2
b.
t
c.
t
d.
3
2
t
1
2
e. t0 (i.e. kinetic energy is constant)
* C.
KE  12 mv 2
v
So we need to find velocity 
dx
dt

d 32
3 1
t  t2
dt
2
2
1
1 3 
9
So : KE  m  t 2   mt
2 2 
8
Thus, KE is proportion al to t.
10. A pendulum consists of a ball of mass m suspended at the end of a massless cord of
length L as shown below. The pendulum is drawn aside through an angle of 60 degrees
with the vertical and released. At the low point of its swing, the speed of the pendulum
ball is:
a.
gL
b.
2 gL
c.
gL
2
d. gL
e. 2gL
*A.
11. A 10 kg body is constrained to move along the X-axis. The potential energy U of the
body in Joules is given as a function of its position X in meters by:
U ( x)  6x 2 - 4x  3
The force on the particle at x = 3 meters is:
a. 32 N in the +X direction.
b. 32 N in the –X direction.
c. 45 N in the +X direction.
d. 45 N in the –X direction.
e. 98 N in th +X direction.
* B. Remember: you will never simply plug and chug into a function---you will take the
integral or the derivative:
F 
dU
 - 12x - 4  4 - 12x
dx
F(3)  - 32 N
12. (1974) A ball of mass m at one end of a string of length R rotates in a vertical circle
just fast enough to keep the string from going slack when the ball is at the top of the
circle. The ball’s speed when it is at the bottom of the circle is:
a.
2 gR
b.
3gR
c.
4 gR
d.
5gR
e.
7 gR
*D. This is a two-step problem. Use the information given to find the speed at the top,
then conserve energy to find the speed at the bottom. DRAW FBD’s!!!
You will get choice “C” if you assume the speed is zero at the top. This is not true—the
rope would have gone slack before the ball reached the top unless it has the speed found
above at the top.
13. (1993) A conservative force has the potential energy function U(x), shown by the
graph above. A particle moving n one dimension under the influence of this force has 1.0
Joules of KE when it is at position X1. Which of the following is a correct statement
about the motion of the particle?
a. It oscillates with maximum position X2 and minimum position X0.
b. It moves to the right of X3 and does not return.
c. It moves to the left of X0 and does not return.
d. It comes to rest at either X0 or X2.
e. It cannot reach either X0 or X2.
*E. Don’t rush through this. The graph tells you the potential energy of the particle. The
problem tells you the KE of the particle. You must add these two pieces of information to
find the TME:
TME  U ( X 1 )  TME( X 1 )  (-2J)  (1J)  - 1 J
Since the particle needs at least ZERO TME to make it to Xo or X2, it will oscillate back
and forth in the gravitational well, but never reach either of these points without
additional energy being added to the system.
Questions 14 and 15:
A ball swings freely back and forth in an arc from point I to point IV, as shown above.
Point II is the lowest point in the path, III is located 0.5 meter above II, and IV is 1 meter
above II. Air resistance is negligible.
14. If the potential energy is zero at point II, where will the kinetic and potential
energies of the ball be equal?
a. At point II.
b. At some point between II and III.
c. At point III.
d. At some point between III and IV.
e. At point IV.
* C. Because gravitational potential energy is linear with height (varies directly), it has
half its max. value at its halfway vertical point.
15. The speed of the ball at point II is most nearly:
a. 3.0 m/s.
b. 4.5 m/s.
c. 9.8 m/s.
d. 14 m/s.
e. 20 m/s.
*B. Conserve energy, recognizing the KE is zero at turnarounds and Ug is zero at lowest
point.
TME IV  KE IV  TME II  KE II
mgh  0  0  12 mv 2
(mass cancels)
v  2gh  2(10m / s 2 )(1m)  4.47m / s
16. (1984) A system consists of two objects having masses
M1 and M2 (M1 < M2). The objects are connected by a
massless string, hung over a pulley as shown above, and then
released. When the object of mass M2 has descended a
distance h, the potential energy of the system has decreased
by:
a. (M2 – M1) gh
b. M2gh
c. (M1 + M2) gh
d. ½ (M1 + M2) gh
e. 0
*A. M2 has LOST gravitational potential energy: M2gh
M1 has GAINED gravitational potential energy of M1gh.
So, as a whole, the SYSTEM has a CHANGE in Ug of: M1gh – M2gh.
However, because the problem says that the Ug has DECREASED, it implies the
negative sign and we want choice A.