HW05 - energy conservation: in which we solve 8P6, 8P24, 8P27, 8P30, 8P38, 8P42, 8P44, 8P48, 8P53, 8P57. •• Chapter 8, problem 6 (|| 8.8): In Fig. 8-31, a small block of mass m = 0.032 kg can slide along the frictionless loop-the-loop, with loop radius R = 12 cm. The block is released from rest at point P, at height h = 5.0R above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point P to (a) point Q and (b) the top of the loop? If the gravitational potential energy of the block – block – Earth Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point P, (d) at point Q, and (e) at the top of the loop? (f ) If,
instead of merely being released, the block is given some initial speed down-ward along the track, do the answers to (a) through (e) increase, decrease, or remain the same?
Solution: We use Eq. 7-12 for W g and Eq. 8-9 for U .
(a) The displacement between the initial point and Q has a vertical component of h – R downward R downward (same
direction as F g ), so (with h = 5 R) R) we obtain,
W g Fg d mg y d2 d1 mg ( R 5 . 0R) mg 4R 4(3 . 2 0 1 02 kg ) (9 . 8 0 sm2 ) (0 . 1 2m) 0 . 1 5 J ; (1.1) ˆ
(b) The displacement between the initial point and the top of the loop has a vertical component of h – 2 2 R
downward (same direction as F g ), so (with h = 5 R) R) in exactly the same manner as (1.1) we obtain,
W g Fg d mg (5.0R 2.0R) 3(3.20 102 kg)(9.80 m/s2 )(0.12 m) 0.11 J ;
(1.2)
(c) With y With y = = h = 5 R, R, at P at P we we find U 5mgR 5(3.20 10 2 kg)(9.80 m/s 2 )(0.12 m) 0.19 J .
(d) With y With y = = R R,, at Q we have U mgR (3.20 102 kg)(9.80 m/s 2 )(0.12 m) 0.038 J .
(e) With y With y = = 2 R, R, at the top of the loop, we find U 2mgR 2(3.20 10 2 kg)(9.80 m/s 2 )(0.12 m) 0.075 J .
(f) The new information (vi 0) is not involved in any of the preceding computations; the above results are unchanged.
•• chapter 8, problem 24: A block of mass m = 2.0 kg is dropped from height h = 40 cm onto a spring of spring constant k = 1960 N/m (picture on left). Find the maximum distance the spring is compressed.
Let down be positive, and let y 0 be the fully-compressed spring-length, implying that the block is dropped from yi h x . The block is of mass m , dropped from h above the spring’s relaxed length, and the spring is subsequently compressed a (positive) displacement x . Consequently, the block drops a total (positive) distance
h x . Conservation of energy, Ei E f , yields, Ei E f Ki U i K f U f U i U f mg h x 12 k x f 2 xi2 12 kx2 2
write in the form Ax Bx C 0 to use quadratic formula
mg h x kx 0; 1 2
(1.3)
2
Using the quadratic formula, (1.3) implies, x
2 ) mg (mg )2 2mghk +-root (19.6 N ) (19.6 N ) 2(19.6 N )(0.40 m)(1960 N m
k
1960 N m
0.10m ;
(1.4)
•• chapter 8, problem 27 (|| 8.7, 8.18, 8.21) : Tarzan, who weighs 688 N, swings from a cliff at the end of a vine 18 m long (Fig. 8-38). From the top of the cliff to the bottom of the swing, he descends by 3.2 m. The vine will break if the force on it exceeds 950 N. (a) Does the vine break? (b) If no, what is the greatest force on it during the swing? If yes, at what angle with the vertical does it break?
Solution: (a) The maximum tension occurs at the lowest point in the swing. Choosing upward positive, and
computing the velocity using K U 12 mv2 mgh , Newton s second law leads to, '
v2 v 2 gh 2 gh 3.20m F ma 0 T mg Fr T m g r m g r 688N 1 2 18.0m 933N (1.5) solve for T
Thus, the vine does not break. (b) Rounding to an appropriate number of significant figures, we see the 2 maximum tension is roughly 9.3 10 N.
•• chapter 8, problem 30 (|| 8.29): A 2.0 kg
breadbox on a frictionless incline of angle 0
40°
acceleration at the instant the box momentarily stops?
is connected, by a cord that runs over a pulley, to a light spring of spring constant k = 120 N/m, as shown in Fig. 8-41. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10 cm down the incline? (b) How far down the incline from its point of release does the box slide before momen-tarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box’s Solution: We take the original height of the box to be the y = 0 reference level and observe that, in general, the height of the box (when the box has moved a distance d downhill) is y d sin 40 .
(a) Using the conservation of energy, we have Ki Ui K U 0 0
1 2
mv2 mgy
1 2
kd 2 .
Therefore, with d = 0.10 m, we obtain v = 0.81 m/s. (b) We look for a value of d 0 such that K = 0. Ki U i K U 0 0 0 mgy
Thus, we obtain mgd sin40
1 2
1 2
2
kd .
kd 2 and find d = 0.21 m.
(c) The uphill force is caused by the spring (Hooke's law) and has magnitude kd = 25.2 N. The downhill force is the component of gravity mg sin 40 = 12.6 N. Thus, the net force on the box is (25.2 – 12.6) N = 12.6 N uphill, with 2
a = F /m =(12.6 N)/(2.0 kg) = 6.3 m/s . (d) The acceleration is up the incline.
•• chapter 8, problem 38 (|| 8.39): Figure 8-47 shows a plot of potential energy U versus position x of a 0.200 kg particle that can travel only along an x-axis under the influence of a conservative force.
The graph has these values: U A
U C 20.00 J and U D
24.00 J .
9.00 J ,
The particle is
released at the point where U forms a “potential hill” of “height” U B
12.00
J , with kinetic energy
4.00 J. What is the speed of the particle at (a) m and (b) m? What is the position of the turning point on (c) the right side and (d) the left side? Solution: In this problem, the mechanical energy (the sum of K and U ) remains constant as the particle moves.
(a) Since mechanical energy is conserved, U B
KB
U A K A , the kinetic energy of the particle in region A (
3.00 m x 4.00 m ) is
K A
U B U A K B
12.0
J 9.00 J 4.00 J
7.00 J
.
2 With K A mvA / 2, the speed of the particle at x 3.5 m (within region A) is
v A
(b) At x 6.5 m, U 0
and K
U B
2 K A m
K B
2(7.00 J)
0.200 kg
12.0
J 4.00 J
8.37 m/s.
16.0
J by
Therefore, the speed at this point is v
2 K m
2(16.0 J) 0.200 kg
(c) At the turning point, the speed of the particle is zero. Let the position of the right turning point be x R . From the figure shown on the right, we find x R to be 16.00 J 0 x R 7.00 m
24.00 J 16.00 J 8.00 m xR
xR 7.67 m.
12.6 m/s.
mechanical energy conservation.
(d) Let the position of the left turning point be x L . From the figure shown, we find x L to be 16.00 J 20.00 J x L 1.00 m
9.00 J 16.00 J
xL 1.73 m.
3.00 m x L
• chapter 8, problem 42: A worker pushed a 27 kg block 9.2 m along a level floor at constant speed with a force directed 32° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.20, what were (a) the work done by the worker’s force and (b) the increase in thermal energy of the block – floor
system? The work done by the worker’s force, from
F
Wa Fa d
Fa d
x
0 Fa cos f (constant speed), is, Fa d
cos Fad f
d ;
(1.6)
In which: the force of friction f is given by,
F
y
0 N mg Fa sin
Fa d
f K N K mg Fa sin
Fa d
mg f sin ; K
Fad
(1.7)
Isolating f in (1.7) yields,
f f K mg f sin Fad solve for f
K mg ; 1 K sin Fad
(1.8)
Putting (1.8) back into (1.6), we finally compute W a Wa
K mg d 1 K sin Fad
(0.20)(27kg )(9.81 sm2 )(9.2m) 1 0.20sin 32
545J ;
(1.9)
(b) Since f k = k (mg + F sin ), we find Eth W a = 545J . Chapter 8, problem 44 (|| 45): A horizontal force of magnitude 35.0 N pushes a block of mass 4.00 kg across a floor where the coefficient of kinetic friction is 0.600. (a) How much work is done by that applied force on the block – floor system when the block slides through a displacement of 3.00 m across the floor?
Work is force dotted into distance,
W F d
F d
cos Fd 35.0 N 3.00 m cos 0 105 J
(1.10)
(b) During that displacement, the thermal energy of the block increases by 40.0 J. What is the increase in 1 thermal energy of the floor? The total amount of energy that has gone to thermal forms is,
Eth Eth,floor Eth,block k mgd 0.600 4.00 kg 9.80 sm 3.00 m 70.6 J Eth,floor 40 J ; (1.11) 2
1
Note: this is independent of whether or not the block-velocity is constant, due to
F 0 F f . a
If 40.0 J has gone to the block then (70.6 – 40.0) J = 30.6 J has gone to the floor. (c) What is the increase in the kinetic energy of the block? Much of the work (105 J) has been “wasted” due to the 70.6 J of thermal energy generated, but there still remains (105 – 70.6 ) J = 34.4 J that has gone into increasing the kinetic energy of the block. (It has not gone into increasing the potential energy of the block because the floor is presumed to be horizontal.) Chapter 8, problem 48 (|| 49): In Fig. 8-49, a block slides down an incline. As it moves from point A to point B, which are 5.0 m apart, force acts on the block, with magnitude 2.0 N and directed down the incline. The magnitude of the frictional force acting on the block is 10 N. If the kinetic energy of the block increases by 35 J between A and B, how much work is done on the block by the gravitational force as the block moves from A to B?
We use Eq. 8-31 to obtain
Eth f k d (10 N)(5.0 m) 50 J and Eq. 7-8 to get W Fd (2.0N)(5.0m) 10 J.
Similarly, Eq. 8-31 gives W K U Eth 10 35 U 50
which yields U = – 75 J. By Eq. 8-1, then, the work done by gravity is W = – U = 75 J. Chapter 8, problem 53 (|| 52): In Fig. 8-50, a 3.5 kg block is accelerated from rest by a precompressed spring of spring constant 640 N/m. The block leaves the spring at the spring’s relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction mu_k = 0.25. The frictional force stops the block in distance D = 7.8 m.
(a) Compute the increase in the thermal energy of the block – floor system. (a) thermal energy is the absolute value of the work done by friction. We have
F
y
0 FN mg , and that
gives a normal force of F N mg f / K ; as indicated, this lets us compute the friction force. Work, then, is,
g g
. kg 9.8 m s W f fD K FN D K ( mg ) D K mgD E th 0.25 35
2
g
7.8 m 67 J. (1.12)
(b) the maximum kinetic energy of the block. The block has its maximum kinetic energy K max just as it leaves the spring and enters the region where friction acts. This kin etic energy equals the work done b y friction, by conservation of energy. This is justified by noting the “boring” fact that the final velocity is 0 m/s, and consequently the final kinetic energy is 0; thus, the maximum kinetic energy equals the initial kinetic energy,
v f 0 Kmax Ki W f 67 J 67 J
(1.13)
(c) the original compression distance of the spring? The energy that appears as kinetic energy is originally in the form of potential energy in the compressed spring. Thus, Kmax Ui
1 2
solve for x kx 2 x
2 K max k
2 67 J
g 0.46 m.
640 N m
Chapter 8, problem 57: a block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed v0 is 6.0 m/s, the height difference h is 1.1 m, and the µk is 0.60. Find d. E U mg y mgh Ki WFr 12 mv0 2 k mgd
d solve for d
mgh 12 mv0 2
k mg
1 2
(v0 2 / g ) h
k
1 2
((6.0 sm ) / 9.81 sm2 ) 1.1m 2
0.60
1.2m
(1.14)
