Problem#1
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 the cannon 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?
Answer:
(a) we use
Wnc = ∆K + ∆U
f∆xcos1800 = Kf – Ki + Uf – Ui
–f∆x = ½ mv2 – 0 + 0 – ½ kx2
–(0.032 0 N)(0.150 m) = ½ (5.30 x 10-3 kg)v2 – ½ (8.00 N/m)(5.00 x 10-2 m)2
5.20 x 10-3 kgm2/s2 = ½ (5.30 x 10-3 kg)v2
v = 1.40 m/s
(b) When the spring force just equals the friction force, the ball will stop speeding up. Here Fs
= kx; the spring is compressed by
x = F/k = (3.20 x 10-2 N)/(8.00 N/m) = 0.400 cm
and the ball has moved 5.00 cm – 0.400 cm = 4.60 cm
(c) Between start and maximum speed points,
Wnc = ∆K + ∆U
f∆xcos1800 = Kf – Ki + Uf – Ui
–f∆x = ½ mv2 – 0 + 0 + ½ kxf2 – ½ kxi2
–(0.032 0 N)(4.60 x 10-2 m) = ½ (5.30 x 10-3 kg)v2 + ½ (8.00 N/m)(4.00 x 10-3 m)2 – ½ (8.00 N/m)(5.00 x 10-2 m)2
1.69 x 10-2 kgm2/s2 = (5.30 x 10-3 kg)v2
v = 1.79 m/s
Problem#2
A 50.0-kg block and a 100-kg block are connected by a string as in Figure 1. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 50.0 kg block and incline is 0.250. Determine the change in the kinetic energy of the 50.0-kg block as it moves from A to B, a distance of 20.0 m.
 |
| Fig.1 |
Answer:
(a) we use
Wnc = (∆K + ∆U)A + (∆K + ∆U)B
f∆xcos1800 = (∆K + ∆U)A + (∆K + ∆U)B
with
n – mgcos37.00 = 0
n = mgcos37.00 = 400 N, and
f = µN = 0.250(400 N) = 100 N
∆UA = mAg(hf – hi) = (50.0 kg)(9.80 m/s2)(20.0 msin37.00) = 5.90 x 103 J
∆UB = mBg(hf – hi) = (100 kg)(9.80 m/s2)(–20.0 m) = –1.96 x 104 J
∆KA = ½ mA(vf2 – vi2) and ∆KB = ½ mB(vf2 – vi2), then ∆KB = (mB/mA)∆KA = 2∆KA
So that
(100 N)(20.0 m)(–1) = 3∆KA + 5.90 x 103 J –1.96 x 104 J
11700 J = 3∆KA
∆KA = 3.90 kJ
Problem#3
A 1.50-kg object is held 1.20 m above a relaxed massless vertical spring with a force constant of 320 N/m. The object is dropped onto the spring. (a) How far does it compress the spring? (b) What If? How far does it compress the spring if the same experiment is performed on the Moon, where g = 1.63 m/s2? (c) What If? Repeat part (a), but this time assume a constant air-resistance force of
0.700 N acts on the object during its motion.
Answer:
(a) The object moved down distance 1.20m + x. Choose y = 0 at its lower point.
Wnc = ∆K + ∆U, with Wnc = 0
0 = Kf – Ki + Ufg + Ufs – Uig – Uis
0 = 0 – 0 + 0 + ½ kx2 – mg(1.20 m + x) – 0
(1.50 kg)(9.80 m/s2)(1.20 m + x) = ½ (320 N/m)x2
(160 N/m)x2 – (14.7 N)x – 17.6 J = 0
x = {14.7 ± [(–14.7)2 – 4(160)(17.6)]1/2}/[2(160)]
x = 0.381 m
(b) From (a) the same equation,
(1.50 kg)(1.63 m/s2)(1.20 m + x) = ½ (320 N/m)x2
(160 N/m)x2 – (2.44 N)x – 2.93 J = 0
x = {2.44 ± [(2.44)2 – 4(160)(2.93)]1/2}/[2(160)]
x = 0.143 m
(c) The equation expressing the energy version of the nonisolated system model has one more term:
Wnc = ∆K + ∆U
–fd = Kf – Ki + Ufg + Ufs – Uig – Uis
–(0.700 N)(1.20 m + x) = 0 – 0 + 0 + ½ kx2 – mg(1.20 m + x) – 0
(1.50 kg)(9.80 m/s2)(1.20 m + x) = ½ (320 N/m)x2 + (0.700 N)(1.20 m + x)
(160 N/m)x2 – (14 N)x – 16.8 J = 0
x = {14.0 ± [(–14.0)2 – 4(160)(16.8)]1/2}/[2(160)]
x = 0.371 m
Problem#4
A 75.0-kg skysurfer is falling straight down with terminal speed 60.0 m/s. Determine the rate at which the skysurfer–Earth system is losing mechanical energy.
Answer:
The total mechanical energy of the skysurfer-Earth system is
Emech = K + Ug = ½ mv2 + mgh
Since the skysurfer has constant speed,
dEmech/dt = mvdv/dt + mgdh/dt = 0 + mg(–v) = –mgv
The rate the system is losing mechanical energy is then
│dEmech/dt│ = (75.0 kg)(9.80 m/s2)(60.0 m/s) = 41.1 kW
Problem#5
A uniform board of length L is sliding along a smooth (frictionless) horizontal plane as in Figure 3a. 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 )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 3b. Find the initial speed v of the board.
 |
| Fig.3 |
(a) Let m be the mass of the whole board. The portion on the rough surface has mass mx/L. The
normal force supporting it is mgx/L and the frictional force is
µkmgx/L = ma
a = µkgx/L opposite to the motion
(b) In an incremental bit of forward motion dx, the kinetic energy converted into internal energy is
fkdx = (µkmgx/L)dx
The whole energy converted is
∫0L(µkmgx/L)dx = ½ mv2
µkmgL/2 = ½ mv2
v = (µkgL)1/2
Post a Comment for "Changes in Mechanical Energy for Nonconservative Forces Problems and Solutions 3"