Download Homework #2 (Ch. 6 – 9) [Solution] 2015 Summer Physics 1

Homework #2 (Ch. 6 ‒ 9) [Solution]
2015 Summer Physics 1
#1.
As shown in the figure, a ball (mass of m) is tied to a ceiling by a
massless string (length of L) and is rotating around with a constant
speed. The angle is θ between the string and the rotational axis.
Answer the following questions.
(a) Find (draw) all the forces acting on the ball.
Look at the right figure.
(b) Express the acceleration of the ball with given parameters.
The perpendicular component of tension is the same of the gravity.
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝜃𝜃 = 𝑚𝑚𝑚𝑚
A ball rotates by the holizontal component of tension which is a
centrifugal force.
𝑇𝑇 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚 tan 𝜃𝜃 = ma𝑐𝑐
Then centrifugal acceleration is
ac = g tan 𝜃𝜃
(c) If m = 2 kg, L = 1 m, and θ = 30o, what is the tension of the string and what is the
acceleration of the ball?
𝑚𝑚𝑚𝑚
(2 𝑘𝑘𝑘𝑘) ∗ ( 9.8 𝑚𝑚/ 𝑠𝑠 2 )
𝑇𝑇 =
=
≅ 22.63 𝑁𝑁
cos 𝜃𝜃
cos 𝜋𝜋/6
𝑎𝑎𝑐𝑐 = 𝑔𝑔 tan 𝜃𝜃 ≃ 5.66 𝑚𝑚/𝑠𝑠 2
(d) Assume the height of the ceiling from the ground is 4 meter. Then, if the string is cut
suddenly and the ball is released, how far in the horizontal direction will the ball fly before it
hits the ground? And how long does it take before it hits the ground?
When the ball hits the ground, the distance from initial position is
1
𝑑𝑑 − 𝐿𝐿 cos 𝜃𝜃 = 𝑔𝑔𝑡𝑡 2
2
Then the time for a free fall is
2(𝑑𝑑 − 𝐿𝐿 cos 𝜃𝜃)
≅ 0.80 𝑠𝑠𝑠𝑠𝑠𝑠
𝑔𝑔
𝑡𝑡 = �
We can find the horizontal component of velocity of the ball from the centrifugal acceleration.
𝑣𝑣 2
𝑣𝑣 2
ac =
=
𝑟𝑟
𝐿𝐿 sin 𝜃𝜃
From (b), the velocity is
𝑣𝑣 = �𝐿𝐿𝐿𝐿 sin 𝜃𝜃 tan 𝜃𝜃 ≅ 1.68 𝑚𝑚/𝑠𝑠
Then the horizontal distance is
vt ≅ 1.34 𝑚𝑚
1
#2.
Two blocks are connected with massless cord, as shown in
figure. Mass of block A and block B is mA = 4 kg and mb = 2 kg,
respectively. The coefficient of kinetic friction between block B
and the horizontal plane is µk = 0.5. The inclined plane is
frictionless and angle θ is 30o. The pulley only changes the
direction of the cord connecting to blocks.
Find (a) the tension of the cord and (b) the magnitude of the
acceleration of the blocks.
(a) For block A, Newton’s law gives
𝑚𝑚𝐴𝐴 𝑔𝑔 sin 𝜃𝜃 − 𝑇𝑇 = 𝑚𝑚𝐴𝐴 𝑎𝑎
where T is tension. For block B, Newton’s law gives
𝑇𝑇 − 𝑓𝑓 = 𝑚𝑚𝐵𝐵 𝑎𝑎
where 𝑓𝑓 = 𝑚𝑚𝐵𝐵 𝑔𝑔𝜇𝜇𝑘𝑘 is the frictional force of the block B. We obtain tension from the above
equations.
𝑚𝑚𝐴𝐴
𝑚𝑚𝐴𝐴 𝑔𝑔 sin 𝜃𝜃 − 𝑇𝑇 =
(𝑇𝑇 − 𝑓𝑓)
𝑚𝑚𝐵𝐵
𝑚𝑚𝐴𝐴 𝑚𝑚𝐵𝐵
(sin 𝜃𝜃 + 𝜇𝜇𝑘𝑘 )g ≅ 13.06 𝑁𝑁
𝑇𝑇 =
𝑚𝑚𝐴𝐴 + 𝑚𝑚𝐵𝐵
(b) In the same way,
𝑎𝑎 =
𝑇𝑇 − 𝑓𝑓 m𝑎𝑎 sin 𝜃𝜃 − 𝑚𝑚𝐵𝐵 𝜇𝜇𝑘𝑘
=
g ≅ 1.63 𝑚𝑚/𝑠𝑠 2
𝑚𝑚𝐵𝐵
𝑚𝑚𝐴𝐴 + 𝑚𝑚𝐵𝐵
2
#3.
Consider that an object with mass of 1 kg slides up along a
frictionless ramp (angle θ = 15o) by a tension T = 10 N. If the
object moves by 10 cm, (a) what is the work done by the
gravitational force and (b) what is the work done by the tension?
(a) The work done by the gravity is
W𝑔𝑔 = − m𝑎𝑎 gsin 𝜃𝜃 ∗ 𝐿𝐿 = −(1 𝑘𝑘𝑘𝑘)(9.8 𝑚𝑚/ 𝑠𝑠 2 )(sin 15o ) (0.1 𝑚𝑚)
≅ −0.25 𝐽𝐽
(b) The work done by the tension is
W𝑇𝑇 = 𝑇𝑇 ∗ 𝐿𝐿 = 1 𝐽𝐽
(c) What will be the speed of the object at the final position (the position that it moves by 10
cm from the initially stationary position) ?
Let us consider work-kinetic energy theorem.
1
𝑚𝑚𝑣𝑣 2 − 0 = 𝑊𝑊𝑇𝑇 + 𝑊𝑊𝑔𝑔
2
Then
𝑣𝑣 = �2(𝑊𝑊𝑇𝑇 + 𝑊𝑊𝑔𝑔 )/𝑚𝑚 ≅ 1.22 𝑚𝑚/𝑠𝑠
(d) Let’s assume there is a friction between the object and ramp surface (the kinetic frictional
coefficient is µk = 0.2), then what will be the speed of the object at the final position (the
position that it moves by 10 cm from the initially stationary position) ?
The work done by the frictional force is
W𝑓𝑓 = −𝜇𝜇𝑘𝑘 𝑚𝑚𝑚𝑚 cos 𝜃𝜃 ∗ 𝐿𝐿 ≅ −0.19 𝐽𝐽
Similarly with (c),
Then
1
𝑚𝑚𝑣𝑣 2 = 𝑊𝑊𝑇𝑇 + 𝑊𝑊𝑔𝑔 + 𝑊𝑊𝑓𝑓
2
𝑣𝑣 = �2(𝑊𝑊𝑇𝑇 + 𝑊𝑊𝑔𝑔 + 𝑊𝑊𝑓𝑓 )/𝑚𝑚 ≅ 1.06 𝑚𝑚/𝑠𝑠
3
#4.
A 0.3 kg block is dropped onto a relaxed vertical spring that has a
spring constant of k = 200 N/m. The block becomes attached to the
spring and compresses the spring 10 cm before stopping. While the
spring is being compressed, what work is done on the block by (a) the
gravitational force on it and (b) the spring force?
(a) The compression of the spring is d = 0.10 m. The work done by the
force of gravity (acting on the block) is
W𝑔𝑔 = 𝑚𝑚𝑚𝑚𝑚𝑚 = (0.3 𝑘𝑘𝑘𝑘)(9.8 𝑚𝑚/𝑠𝑠 2 )(0.1 𝑚𝑚) ≅ 0.29 𝐽𝐽
(b) The work done by the spring is
1
W𝑘𝑘 = − 𝑘𝑘𝑑𝑑 2 = −(0.5)(200 𝑁𝑁/𝑚𝑚 )(0.1𝑚𝑚)2 = −1𝐽𝐽
2
(c) What is the speed of the block just before it hits the spring?
Let us consider work-kinetic energy theorem.
1
0 − mvi 2 = 𝑊𝑊𝑔𝑔 + 𝑊𝑊𝑘𝑘
2
Then the speed of the block just before it hits the spring is
vi = �−2(𝑊𝑊𝑔𝑔 + 𝑊𝑊𝑘𝑘 )/𝑚𝑚 ≅ 2.17 𝑚𝑚/𝑠𝑠
(d) If the speed at impact is doubled, what is the maximum compression of the spring?
If the speed of the block just before it hits the spring is given as v, we obtain new length of
compression of spring d′. Recall work-kinetic energy theorem.
1
1
0 − mv 2 = 𝑊𝑊𝑔𝑔 + 𝑊𝑊𝑘𝑘 = 𝑚𝑚𝑚𝑚𝑑𝑑′ − 𝑘𝑘𝑑𝑑′2
2
2
Then
𝑚𝑚𝑚𝑚 + �𝑚𝑚2 g 2 + mkv 2
𝑑𝑑′ =
k
If v = 2vI ,
d′ ≅ 0.18 𝑚𝑚
4
#5.
A person (weight of 60 kg) is doing bungee jump from a bridge at
45 meter above the river surface (H = 45 m). Bungee cable is 25
meter long (L = 25 m) and its elastic coefficient is 150 N/m.
(a) When the bungee jumper reaches a stop after bungee jump, what
would be the height h?
The difference of the potential energy of gravity is
Vg = −mg(L + d)
The elastic potential energy of bungee cable is
1
V𝑒𝑒 = 𝑘𝑘𝑑𝑑 2
2
where k is elastic coefficient. Then let us consider conservation of
total energy.
𝑉𝑉𝑔𝑔 + 𝑉𝑉𝑒𝑒 = 0
Then
𝑚𝑚𝑚𝑚 + �𝑚𝑚2 g 2 + 2m𝑘𝑘𝑘𝑘𝑘𝑘
≅ 18.46 𝑚𝑚
𝑑𝑑 =
k
h = H − L − d ≅ 1.54 𝑚𝑚
(b) At the position of problem (a), what is the net force acting on the bungee jumper?
The net force is
F = −mg + kd ≅ 2181 𝑁𝑁
Where the direction of net force is upward.
5
L
ㅣ
H
d
h
#6.
A single conservative force F(x) acts on a 1 kg particle that moves along an axis. The
potential energy U(x) associated with F(x) is given by
U(x) = −4x e −2x/5 J
where x is in meters. At x = 5 m, the particle has a kinetic energy of 1 J.
(a) What is the mechanical energy of the system?
E = U(5) + 𝑇𝑇(5) ≅ −2.71 𝐽𝐽 + 1 𝐽𝐽 = −1.71 𝐽𝐽
Where T(5) is a kinetic energy at x = 5.
(b) Make a plot of U(x) as a function of x for 0 ≤ x ≤ 10 m, and on the same graph draw the
line that represents the mechanical energy of the system.
(c) Use part (b) to determine the least value and the greatest value of x the particle can reach.
(Write only equation and indicate the point at (b) graph.)
The least value and the greatest value stratify
Then,
2𝑥𝑥
U(x) = −4𝑥𝑥𝑒𝑒 − 5 = −1.71 𝐽𝐽
x𝑚𝑚𝑚𝑚𝑚𝑚 ≅ 0.53 m, x𝑚𝑚𝑚𝑚𝑚𝑚 ≅ 6.98 𝑚𝑚
(d) Use part (b) to determine the maximum kinetic energy of the particle and the value of x
at which it occurs.
Seeing with the naked eye, the maximum kinetic energy may be 1.9 J. When the kinetic
energy is maximum, the value of x satisfy
Then
2𝑥𝑥
1.9 − 4𝑥𝑥𝑒𝑒 − 5 = −1.71
x ≅ 2.5 ± 0.5 m
(e) Determine an expression in newtons and meters for F(x) as a function of x. For what
finite value of x does F(x) = 0?
The conservative force is
2𝑥𝑥
dU
8
𝐹𝐹(𝑥𝑥) = −
= �4 − 𝑥𝑥� 𝑒𝑒 − 5
𝑑𝑑𝑑𝑑
5
The finite value of x such that F(x) = 0 is 5/2.
6
#7.
A block (mass m = 2 kg) slides to a spring (spring
constant k = 300 N/m). When the block stops, it has
compressed the spring by 7 cm. The coefficient of
kinetic friction between block and floor is 0.25. While
the block is in contact with the spring and being brought
to stop, what are (a) the work done by the spring force
and (b) the increase in thermal energy of the block-floor
system?
(a) The work done by the spring force is
1
Ws = − 𝑘𝑘𝑑𝑑2 = −0.735 𝐽𝐽
2
(b) The work done by the frictional force is
Wf = −𝜇𝜇𝑘𝑘 𝑚𝑚𝑚𝑚𝑚𝑚 = −0.343 𝐽𝐽
The increase in thermal energy is
∆U = −Wf = 0.343 J
(c) What is the block’s speed just as it reaches the spring?
Let us consider work-kinetic energy theorem.
1
0 − 𝑚𝑚𝑣𝑣 2 = 𝑊𝑊𝑠𝑠 + 𝑊𝑊𝑓𝑓
2
Then
v = �−2(𝑊𝑊𝑠𝑠 + 𝑊𝑊𝑓𝑓 )/𝑚𝑚 ≅ 1.04 𝑚𝑚/𝑠𝑠
7
#8.
Find a center of mass of a uniform plate P of radius R
from which a disk of its diameter R has been removed
and attached to the right side of it as shown in the figure.
The center of mass of a uniform plate of radius R without
hole is
P1 = (0, 0)
The center of mass of a small plate of radius R/2 is
P2 = (1.5𝑅𝑅, 0)
The center of mass of a hole is
p3 = (−0.5𝑅𝑅, 0)
Let us consider superposition law of center of mass. The center of mass of whole systems is
𝑅𝑅 2
𝑅𝑅 2
𝜋𝜋𝑅𝑅2 𝑃𝑃1 + 𝜋𝜋 � � 𝑃𝑃2 − 𝜋𝜋 � � 𝑃𝑃3
2
2
P=
= (0.5R, 0)
2
2
𝑅𝑅
𝑅𝑅
2
𝜋𝜋𝑅𝑅 + 𝜋𝜋 � � − 𝜋𝜋 � �
2
2
8
#9.
Block 1 (mass of m1) slides from rest along a frictionless ramp from height h = 4 m and then
collides with stationary block 2 (mass of m2 = twice of mass m1). After the collision, block 2
slides into a region where the coefficient of kinetic friction µk = 0.4 and stops in distance d
within that region. What is the value of distance d if the collision is (a) elastic and
(b) completely inelastic ?
µk
(a) The speed of the block1 just before it hits the block2 is
1
2
2
𝑚𝑚1 𝑣𝑣1𝑖𝑖
= 𝑚𝑚1 𝑔𝑔ℎ, v1i = �2𝑔𝑔ℎ
For the elastic collision, the speed of the block2 is
v2 =
2𝑚𝑚1
2
𝑣𝑣1𝑖𝑖 = �2𝑔𝑔ℎ
𝑚𝑚1 + 𝑚𝑚2
3
Let us consider work-kinetic energy theorem.
Then the distance is
1
0 − 𝑚𝑚2 𝑣𝑣22 = −𝜇𝜇𝑘𝑘 𝑚𝑚2 𝑔𝑔𝑔𝑔
2
d=
𝑣𝑣22
4ℎ
=
≅ 4.44 𝑚𝑚
2𝜇𝜇𝑘𝑘 𝑔𝑔 9𝜇𝜇𝑘𝑘
v2 =
𝑚𝑚1
1
𝑣𝑣1𝑖𝑖 = �2𝑔𝑔ℎ
𝑚𝑚1 + 𝑚𝑚2
3
d=
𝑣𝑣22
ℎ
=
≅ 1.11 𝑚𝑚
2𝜇𝜇𝑘𝑘 𝑔𝑔 9𝜇𝜇𝑘𝑘
(b) For the completely inelastic collision, the speed of the block2 is
Likewise (a), the distance is
9