Download Energy unit review solutions.

1. Two objects are connected by a light string passing over a light frictionless pulley as in Figure P8.13. The object of mass m1 is released from rest at height h. Using the principle of conservation of energy, (a) determine the speed of m2 just as m1 hits the ground. (b) Find the maximum height to which m2 rises. P8.14 m1 > m2
(a)
m1 gh =
v=
(b)
1
(m + m2 )v2 + m2 gh
2 1
2 (m1 − m2 ) gh
(m1 + m2 )
Since m2 has kinetic energy
1
m v2 , it will rise an additional height Δh
2 2
determined from
m2 g Δh =
1
m v2
2 2
or from (a),
Δh =
v2 ( m1 − m2 )h
=
(m1 + m2 )
2g
The total height m2 reaches is h + Δh =
2m1 h
.
m1 + m2
2. A light rigid rod is 77.0 cm long. Its top end is pivoted on a low-­‐‑friction horizontal axle. The rod hangs straight down at rest, with a small massive ball attached to its bottom end. You strike the ball, suddenly giving it a horizontal velocity so that it swings around in a full circle. What minimum speed at the bottom is required to make the ball go over the top of the circle? P8.15 The force of tension and subsequent force of compression in
the rod do no work on the ball, since they are perpendicular to each
step of displacement. Consider energy conservation of the ball-Earth
system between the instant just after you strike the ball and the instant
when it reaches the top. The speed at the top is zero if you hit it just
hard enough to get it there.
initial
L
final
L
K i + U gi = K f + U gf :
1
mv2i + 0 = 0 + mg(2L )
2
vi
vi = 4gL = 4( 9.80) (0.770 )
vi = 5.49 m s
FIG. P8.15
3. A daredevil plans to bungee-­‐‑jump from a balloon 65.0 m above a carnival midway. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10.0 m above the ground. Model his body as a particle, and the cord as having negligible mass and one that obeys Hooke’s force law. In a preliminary test, hanging at rest from a 5.00-­‐‑m length of the cord, he finds that his body weight stretches it by 1.50 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon. (a) What length of cord should he use? (b) What maximum acceleration will he experience? *P8.19
(a)
For a 5-m cord the spring constant is described by F = kx ,
mg = k (1.5 m) . For a longer cord of length L the stretch distance
is longer so the spring constant is smaller in inverse proportion:
5 m mg
= 3.33 mg L
L 1.5 m
K + U g + U s = K + U g + Us
k=
(
) (
i
)
f
1
0 + mgyi + 0 = 0 + mgy f + kx 2f
2
mg 2
1 2 1
mg y i − y f = kx f = 3.33
x
2
2
L f
(
)
here yi − y f = 55 m = L + x f
initial
final
FIG. P8.19(a)
1
2
3.33( 55.0 m − L )
2
55.0 mL = 5.04 × 103 m2 − 183 mL + 1.67L2
55.0 mL =
0 = 1.67L2 − 238L + 5.04 × 103 = 0
L=
(
238 ± 2382 − 4(1.67 ) 5.04 × 103
)
2(1.67 )
=
238 ± 152
= 25.8 m
3.33
only the value of L less than 55 m is physical.
(b)
mg
25.8 m
∑ F = ma
k = 3.33
x max = x f = 55.0 m − 25.8 m = 29.2 m
+kx max − mg = ma
mg
3.33
29.2 m − mg = ma
25.8 m
a = 2.77 g = 27.1 m s 2
4. A particle of mass m = 5.00 kg is released from point A and slides on the frictionless track shown in Figure P8.24. Determine (a) the particle'ʹs speed at points B and C and (b) the net work done by the gravitational force in moving the particle from A to C. P8.24
(a)
( ΔK) A →B = ∑W = W g = mgΔh = mg( 5.00 − 3.20)
1
1
mv2B − mv2A = m(9.80 )(1.80 )
2
2
vB = 5.94 m s
A
B
5.00 m
3.20 m
2
Similarly, vC = v A + 2g( 5.00 − 2.00 ) = 7.67 m s
(b)
Wg
A →C
FIG. P8.24
= mg( 3.00 m) = 147 J
5. A 50.0-­‐‑kg block and 100-­‐‑kg block are connected by a string as in Figure P8.36. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 50-­‐‑kg block and incline is 0.250. Determine the change in the kinetic energy of the 50-­‐‑kg block as it moves from A to B, a distance of 20.0 m C
2.00 m
P8.36
∑ Fy = n − mg cos37.0° = 0
∴ n = mg cos 37.0° = 400 N
f = µn = 0.250( 400 N) = 100 N
−fΔx = ΔE mech
( −100)( 20.0) = ΔUA
+ ΔU B + ΔKA + ΔKB
(
)
= m g( h − h ) = (100)( 9.80)( −20.0) = −1.96 × 10
1
= m (v − v )
2
1
m
= m (v − v ) =
ΔK = 2ΔK
2
m
ΔUA = mA g h f − hi = (50.0 )( 9.80)( 20.0sin 37.0°) = 5.90 × 103
ΔUB
ΔKA
ΔKB
B
f
A
B
4
i
2
f
2
f
2
i
2
i
B
A
A
A
Adding and solving, ΔKA = 3.92 kJ .
FIG. P8.36
6. A uniform board of length L is sliding along a smooth (frictionless) horizontal plane, as in Figure P8.39a. The board then slides across the boundary with a rough horizontal surface. The coefficient of kinetic friction between the board and the second surface is µ k. (a) Find the acceleration of the board at the moment its front end has traveled a distance x beyond the boundary. (b) The board stops at the moment its back end reaches the boundary, as in Figure P8.39b. Find the initial speed v of the board. *P8.39
(a)
Let m be the mass of the whole board. The portion on the rough surface has
mx
mxg
mass
. The normal force supporting it is
and the frictional force is
L
L
µk mgx
µ gx
= ma . Then a = k
opposite to the motion.
L
L
(b)
In an incremental bit of forward motion dx, the kinetic energy converted into
µ mgx
internal energy is fk dx = k
dx . The whole energy converted is
L
L
µ mgx
µ mg x2
1
mv2 = ∫ k
dx = k
2
L
L 2
0
L
=
0
µk mgL
2
v = µk gL
7. A toy cannon uses a spring to project a 5.30-­‐‑g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m. When it is fired, the ball moves 15.0 cm through the horizontal barrel of the cannon, and there is a constant friction force of 0.032 0 N between the barrel and the ball. (a) With what speed does the projectile leave the barrel of the cannon? (b) At what point does the ball have maximum speed? (c) What is this maximum speed? P8.35
(a)
( K + U ) i + ΔEmech = ( K + U) f
0+
:
1 2
1
kx − fΔx = mv2 + 0
2
2
2
1
1
(
8.00 N m) 5.00 × 10−2 m − 3.20 × 10 −2 N ( 0.150 m) = 5.30 × 10 −3 kg v2
2
2
(
v=
(b)
) (
(
2 5.20 × 10−3 J
5.30 × 10
−3
)=
kg
)
(
)
1.40 m s
When the spring force just equals the friction force, the ball will stop speeding
up. Here F s = kx ; the spring is compressed by
3.20 × 10 −2 N
= 0.400 cm
8.00 N m
and the ball has moved
5.00 cm − 0.400 cm = 4.60 cm from the start.
(c)
Between start and maximum speed points,
1 2
1
1
kx i − fΔx = mv2 + kx 2f
2
2
2
1
1
1
−2 2
8.00 5.00 × 10
− 3.20 × 10 −2 4.60 × 10−2 = 5.30 × 10 −3 v2 + 8.00 4.00 × 10 −3
2
2
2
v = 1.79 m s
(
) (
)(
)
(
)
(
)
2
8. The mass of a car is 1 500 kg. The shape of the body is such that its aerodynamic drag coefficient is D = 0.330 and the frontal area is 2.50 m2. Assuming that the drag force is proportional to v2 and neglecting other sources of friction, calculate the power required to maintain a speed of 100 km/h as the car climbs a long hill sloping at 3.20°. P8.50 v = 100 km h = 27.8 m s
The retarding force due to air resistance is
R=
1
1
2
DρAv2 = ( 0.330 ) 1.20 kg m3 2.50 m 2 ( 27.8 m s) = 382 N
2
2
(
)(
)
Comparing the energy of the car at two points along the hill,
K i + U gi + ΔE = K f + U gf
or
K i + U gi + ΔWe − R ( Δs) = K f + U gf
where ΔWe is the work input from the engine. Thus,
(
) (
ΔWe = R ( Δs) + K f − K i + U gf − U gi
)
Recognizing that K f = Ki and dividing by the travel time Δt gives the required
power input from the engine as
# Δy &
# ΔW e &
# Δs &
P =%
( = R % ( + mg%
( = Rv + mgvsin θ
$ Δt '
$ Δt '
$ Δt '
(
)
P = ( 382 N) (27.8 m s) + (1 500 kg) 9.80 m s 2 (27.8 m s ) sin 3.20°
P = 33.4 kW = 44.8 hp
9. A 10.0-­‐‑kg block is released from point A in Figure P8.57. The track is frictionless except for the portion between points B and C , which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2 250 N/m, and compresses the spring 0.300 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between B and C. €
P8.57
ΔEmech = − fΔx
E f − Ei = −f ⋅ dBC
1 2
kx − mgh = −µmgdBC
2
mgh − 21 kx 2
µ=
= 0.328
mgdBC
FIG. P8.57
10. Jane, whose mass is 50.0 kg, needs to swing across a river (having width D) filled with man-­‐‑eating crocodiles to save Tarzan from danger. She must swing into a wind exerting constant horizontal force F, on a vine having length L and initially making an angle θ with the vertical (Fig. P8.65). Taking D = 50.0 m, F = 110 N, L = 40.0 m, and θ = 50.0°, (a) with what minimum speed must Jane begin her swing in order to just make it to the other side? (b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing? Assume that Tarzan has a mass of 80.0 kg. (a)
From takeoff to alighting for the Jane-Earth system
(K + U )
g
i
(
+ Wwind = K + U g
)
f
1
mv2i + mg( −Lcos θ ) + FD( −1) = 0 + mg( −Lcos φ )
2
1
50 kg v2i + 50 kg 9.8 m s2 ( −40 m cos 50°) − 110 N( 50 m)
2
= 50 kg 9.8 m s2 ( −40 m cos28.9° )
(
(
)
)
1
50 kg v2i − 1.26 × 104 J − 5.5 × 103 J = −1.72 × 104 J
2
2 ( 947 J)
vi =
= 6.15 m s
50 kg
(b)
For the swing back
1
mv2i + mg( −Lcos φ ) + FD( +1) = 0 + mg( −Lcos θ )
2
1
130 kg v2i + 130 kg 9.8 m s2 ( −40 m cos28.9° ) + 110 N(50 m)
2
= 130 kg 9.8 m s2 ( −40 mcos 50°)
(
)
(
)
1
130 kg v2i − 4.46 × 104 J + 5 500 J = −3.28 × 104 J
2
vi =
2 (6 340 J)
= 9.87 m s
130 kg
11. A single conservative force acting on a particle varies as F = − Ax + Bx ˆi N
, where A and B are constants and x is in meters. (a) Calculate the potential-­‐‑
energy function U(x) associated with this force, taking U = 0 at x = 0. (b) Find the change in potential energy and the change in kinetic energy as the particle moves from x = 2.00 m to x = 3.00 m. (
x
P8.40
(a)
(
)
U = −∫ −Ax + Bx2 dx =
0
3.00 m
(b)
ΔU = −
∫
Fdx =
[(
A 3.00 2 − ( 2.00)2
)
2
] − B[( 3.00)
3
− ( 2.00)3
3
]=
5.00
19.0
A−
B
2
3
$ 5.00
19.0 '
ΔK = & −
A+
B)
%
2
3 (
12. A potential energy function for a two-­‐‑dimensional force is of the form U = 3x3y – 7x. Find the force that acts at the point (x, y). (
)
(
)
∂ 3x3 y − 7x
∂U
=−
= − 9x2 y − 7 = 7 − 9x2 y
∂x
∂x
∂ 3x3 y − 7x
∂U
Fy = −
=−
= − 3x3 − 0 = −3x 3
∂y
∂y
(
(
)
Ax 2 Bx 3
−
2
3
2.00 m
P8.42 Fx = −
2
)
)
Thus, the force acting at the point (x, y) is F = Fx ˆi + Fy ˆj = (7 − 9x 2 y)ˆi − 3x3 ˆj .