HW04 – energy: energy: 7P9, 7P13, 7P16, 7P22, 7P24, 7P31, 7P37, 7P42(H), 7P44, 7P49 Chapter 7, problem 9: The only force acting on a 2.0 kg canister that is moving in an xy-plane has a magnitude of 5.0 N. T he canister initially has a velocity of 4.0 m/s in the positive x-direction and some time later has a velocity of 6.0 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time? The time? The final answer is insensitive to the magnitude of the force,
W K f K i 12 m f v f 2 12 mivi2 12 m v f 2 v i2 12 (2.0kg ) ( 02 (6.0 ms ) 2 ) 2 ( (4.0 ms ) 2 0 2 ) 2
(2.0kg ) 36 1 2
m2 s 2
16
m2 s2
20 kg
m2 s2
(1.1)
;
•• Chapter 7, problem 13: A luge and its rider, with a total mass of 85 kg, emerge from a downhill track onto a horizontal straight track with an initial speed of 37 m/s. If a force slows them to a stop at a t a constant rate of 2.0 2 m/s , (a) what magnitude F is required for the force, (b) what distance d do they travel while slowing, and (c) 2 what work W is done on them t hem by the force? What are (d) F, (e) d, and (f) W if they, the y, instead, slow at 4.0 m/s ?
x as the direction of motion (so a and F are negative-valued). negative-valued). (a) Newton’s second law Solution: We choose + x as
5kg)( 2.0m/ .0m/s 2 ) so that, readily yields F (85kg)(
F | F | 1.7 102 N
(1.2)
(b) From Eq. 2-16 (with v = 0) we have, solve for x
v 0 v 2ax x 2
2
2 0
v02 2a
37m/s
2
2 2.0 m/s2
3.4 102 m ;
(1.3)
1
Alternate solution: this can be worked using energ y-conservation , as,
K W 12 mv 2 12 mv0 2 F x ma x max
v 2 v0 2 2
ax x
02 v0 2
2a
37m/s
2
2 2.0 m/s 2
3.4 102 m
(1.4)
I strongly recommend you work this problem out both ways. (c) Since F is opposite to the direction of motion (so the angle between F and d x is 180°) then Eq. 7-7
gives the work done as W F x 5.8 104 J . (d) In this case, Newton’s second law yields, F ma 85 kg 4.0 m/s2 x F F 3.4 10 1 02 N ; ˆ
(1.5)
(e) From (1.3) or (1.4), we now have,
x
37m/s
2
2 4.0 m/s
2
1.7 102 m.
(1.6)
(f) The force F is again opposite to the direction of motion (so the angle is is again 180°) so that Eq. 7-7 leads 4 to W F x 5.8 10 J. The fact that this agrees with the result of part (c) provides insight into the concept
of work.
1
Thus, the formula v
2
v0 2 2a x x can be derived from K W 12 mv 2 12 mv0 2 F x ma x .
•• Chapter 7, problem 16: An 8.0 kg object is moving in the positive direction of an x-axis. When it passes through x = 0, a constant force directed along the axis begins to act on it. Figure 7-28 gives its kinetic energy K versus position x as it moves from x = 0 to x = 5.0 m; K 0 = 30.0 J. The force continues to act. What is v when the object moves back through x = -3.0 m?
Solution: The change in kinetic energy is computed to eventually compute a in the same manner as in (1.4). Note that the change in kinetic energy is NEGATIVE; i.e., the object is slowing down.
We expect, therefore, a negative acceleration (since the object was moving in the positive x-direction),
1
1
2
2
solve for a K m(v f2 vi2 ) m(2a x) ma x a
K ( 30 J) m 0.75 2 ; (1.7) m x (8.0 kg)(5.0 m) s
When x 5 m the kinetic energy becomes zero, implying that the mass comes to rest momentarily. Thus, v 2 v0 2 2a x v0
v 2 2a x
0 2( 0.75 m/s 2 )(5.0 m) 7.5 m 2 /s 2 2.7 sm
(1.8)
The speed of the object when x = 3.0 m is,
v v02 2a x 7.5 m2 /s2 2( 0.75 m/s2 )( 3.0 m) 12 m/s 3.5 m/s ;
(1.9)
•• Chapter 7, problem 22: A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of 10.0 m: (a) the initially stationary spelunker is accelerated to a speed of 5.00 m/s; (b) he is then lifted at the constant speed of 5.00 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 80.0 kg rescue-ee by the force lifting him during each stage? Solution: We use d to denote the magnitude of the spelunker’s displacement during each stage. The mass o f the spelunker is m = 80.0 kg. The work done by the lifting force is denoted W i where i = 1, 2, 3 for the three stages. We apply the work-energy theorem, Eq. 7-15, in the form,
K K f Ki j Wj Wa U g Wi app m( g )( di ) Wi app mgd i ; 2 (a) For stage 1, W1 mgd K1 21 mv1 , where v1 5.00 m / s . This gives
W1 mgd
1 2
mv12 (80.0 kg)(9.80 m/s2 )(10.0 m)
1 2
(80.0 kg)(5.00 m/s)2 8.84 103 J.
(b) For stage 2, W 2 – mgd = K 2 = 0, which leads to W2 mgd (80.0 kg)(9.80 m/s 2 )(10.0 m) 7.84 103 J.
(c) For stage 3, W3 mgd K3 21 mv12 . We obtain
(1.10)
W3 mgd
1 2
mv12 (80.0 kg)(9.80 m/s2 )(10.0 m)
•• Chapter 7, problem 24: In Fig. 7-32, a horizontal force
Fa
1 2
(80.0 kg)(5.00 m/s)2 6.84 103 J.
energy at the start of the displacement, what is its speed at the end of the displacement?
of magnitude 20.0 N is applied
to a m = 3.00 kg psychology book as the book slides a distance d = 0.500 m up a frictionless ramp at angle = 30.0°. (a) During the displacement, what is the net work done on the book by
Fa ,
the
gravitational force on the book, and the normal force on the book? (b) If the book has zero kinetic
(a) Find the net work done on the textbook: Using notation common to many vector-capable calculators, we have (from Eq. 7-8),
W
F d F d x
x
Fa cos ma sin d 20.0x cos30.0 (9.81)(3.00)sin 30.0 N 0.500m 1.31J ; (1.11) ˆ
The normal force makes no contribution to the dot product, since the angle between the normal force and the displacement is 90 . (b) Eq. 7-10 (along with Eq. 7-1) then leads to
Fa d cos mgd sin
K W 12 mv2 Fa d cos mgd sin v
1 2
m
2(1.31J ) 3.00kg
0.935
m s
; (1.12)
•• Chapter 7, problem 31: The only force acting on a 2.0 kg body as it moves along a positive x axis has an x component F x 6 x
N m
, with x in meters. The velocity at x = 3.0 m is 8.0 m/s. (a) What is the velocity of the
body at x = 4.0 m? (b) At what positive value of x will the body have a velocity of 5.0 m/s? (a) As the body moves along the x axis from xi = 3.0 m to x f = 4.0 m the work done by the force is
W
x f
xi
F x dx
x f
xi
6x dx 3( x 2f xi2 ) 3 (4.02 3.02 ) 21 J.
According to the work-kinetic energy theorem, this gives the change in the kinetic energy: W K
1 2
md v f2 vi2 i
where vi is the initial velocity (at xi) and v f is the final velocity (at x f ). The theorem yields
v f
2W m
vi2
2(21 J) 2.0 kg
(8.0 m/s)2 6.6 m/s.
(b) The velocity of the particle is v f = 5.0 m/s when it is at x = x f . The work-kinetic energy theorem is used to solve for x f . The net work done on the particle is W 3 x f2 xi2 , so the theorem leads to 1
3d x f2 xi2 i m d v 2f vi2 i. 2
Thus, x f
m
v 6
2 f
vi2 xi2
2.0 kg
(5.0 m/s) 6 N/m
2
(8.0 m/s) 2 (3.0 m) 2 4.7 m.
•• chapter 7, problem 37: Figure 7-39 gives the acceleration of a 2.00 kg particle as an applied force
Fa
moves it from rest along an x-axis from x0 to x = 9.0 m. The scale of the figure’s vertical axis is set by
a s 6.0 sm2 . How much work has the force done on the particle when the particle reaches (a) x = 4.0 m, (b) x = 7.0 m, and (c) x = 9.0 m? What is the particle’s speed and direction of travel when it reaches (d) x = 4.0 m, (e)
x = 7.0 m, and (f) x = 9.0 m? (a) We first multiply the vertical axis by the mass, so that it becomes a graph of the applied force. Now, adding the triangular and rectangular “areas” in the graph (for 0 x 4) gives 42 J for the work done. (b) Counting the “areas” under the axis as negative contributions, we find (for 0 x 7) the work to be 30 J at x = 7.0 m. (c) And at x = 9.0 m, the work is 12 J. (d) Equation 7-10 (along with Eq. 7-1) leads to speed v = 6.5 m/s at x = 4.0 m. Returning to the original graph (where a was plotted) we note that (since it started from rest) it has received acceleration(s) (up to this point) only in the + x direction and consequently must have a velocity vector pointing in the + x direction at x = 4.0 m. (e) Now, using the result of part (b) and Eq . 7-10 (along with Eq. 7-1) we find the speed is 5.5 m/s at x = 7.0 m. Although it has experienced some deceleration during the 0 x 7 interval, its velocity vector still points in the + x direction. (f) Finally, using the result of part (c) and Eq. 7 -10 (along with Eq. 7-1) we find its speed v = 3.5 m/s at x = 9.0 m. It certainly has experienced a significant amount of deceleration during the 0 x 9 interval; nonetheless, its velocity vector still points in the + x direction. Chapter 7, Problem 42: The figure shows a cord attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height h = 1.20 m, so the cart slides from x1 = 3.00 m to x2 = 1.00 m. During the move, the tension in the cord is a
constant 25.0 N. What is the change in the kinetic energy of the cart during the move?
The work-energy theorem is used, but one must use the identity cos
adjacent hypotenuse
upon W F d
F d
cos , and
eliminate the trig-functions in favour of the dimensions illustrated above. In fact, this should not be a 3-dot problem, since this is not a long problem, but rather one which requires a touch of ingenuity, K Wa F (x 2 x1 )
(25.0 N )
F
x
2
cos 2
x1
cos 1 F
(1.00m) (1.20m) (3.00m) (1.20m) 2
2
2
2
x2 2 h 2
x12 h 2
41.7 J 41.7J
(1.13)
• chapter 7, problem 44: A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12° with the horizontal. The rope moves parallel to th e slope with a constant speed of 1.0 m/s. The force of the rope does 900 J of work on the skier as the skier moves a distance of 8.0 m up the incline. (a) If the rope moved with a constant speed of 2.0 m/s, how much work would the force of the rope do on the skier as the skier moved a distance of 8.0 m up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) 1.0 m/s and (c) 2.0 m/s?
(a) Since constant speed implies K 0, we require Wa W g , by Eq. 7-15. Since W g is the same in both 2 cases (same weight and same path), then W a 9.0 10 J just as it was in the first case.
(b) Since the speed of 1.0 m/s is constant, then 8.0 meters is traveled in 8.0 seconds. Using Eq. 7-42, and noting that average power is the power when the work is being done at a steady rate, we have W 900 J 1.1102 W. P t 8.0 s (c) Since the speed of 2.0 m/s is constant, 8.0 meters is traveled in 4.0 secon ds. Using Eq. 7-42, with average power replaced by power , we have W 900 J = 225 W 2.3 102 W . P t 4.0 s •• chapter 7, problem 49: A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 54 m in 3.0 min, starting and ending at rest. The elevator’s counterweight has a mass of only 950 kg, and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable?
We have a loaded elevator moving upward at a constant speed. The forces involved are: gravitational force on the elevator, gravitational force on the counterweight, and the force by the motor via cable. The total work is the sum of the work done by gravity on the elevator, the work done by gravity on the counterweight, and the work done by the motor on the system: W We Wc Wm .
Since the elevator moves at constant velocity, its kinetic energy does not change and according to the workkinetic energy theorem the total work done is zero, that is, W K 0 . The elevator moves upward through 54 m, so the work done by gravity on it is
We me gd (1200 kg)(9.80 m/s2 )(54 m) 6.35 105 J.
The counterweight moves downward the same distance, so the work done by gravity on it is Wc mc gd (950 kg)(9.80 m/s 2 )(54 m) 5.03 105 J.
Since W = 0, the work done by the motor on the system is Wm We W c 6.35 105 J 5.03105 J 1.32 105 J.
This work is done in a time interval of t 3.0 min 180 s, so the power supplied by the motor to lift the elevator is P
W m
t
1.32 105 J 180 s
7.4 102 W.
