Problem#1
A single constant force F acts on a particle of mass m. The particle starts at rest at t = 0. (a) Show that the instantaneous power delivered by the force at any time t is P = (F2/m)t. (b) If F = 20.0 N and
m = 5.00 kg, what is the power delivered at t = 3.00 s?Answer:
The power done by force F is given by
P = F.v
with v = vi + at or v = 0 + Ft/m, then
P = F2t/m
(b) If F = 20.0 N and m = 5.00 kg, the power delivered at t = 3.00 s is
P = (20.0 N)2(3.00 s)/(5.00 kg) = 240 W
Problem#2
Two springs with negligible masses, one with spring constant k1 and the other with spring constant k2, are attached to the endstops of a level air track as in Figure 1. A glider attached to both springs is located between them. When the glider is in equilibrium, spring 1 is stretched by extension xi1 to the right of its unstretched length and spring 2 is stretched by xi2 to the left. Now a horizontal force F app is applied to the glider to move it a distance xa to the right from its equilibrium position. Show that in this process (a) the work done on spring 1 is ½ k1(xa2 + 2xaxi1), (b) the work done on spring 2 is ½ k2(xa2 – 2xaxi2), (c) xi2 is related to xi1 by xi2 = k1xi1/k2, and (d) the total work done by the force Fapp is ½ (k1 + k2)xa2.
Answer:
(a) the work done on spring 1 is ½ k1(xa2 + 2xaxi1),
W1 = ∫ifF1dx = ∫xi1(xi1 + xa)k1xdx
= ½ k1x2│xi1(xi1 + xa)
= ½k1 [(xi1 + xa)2 – xi12]
W1 = ½k1[(xi1 + xa)2 – xi12]
W1 = ½ k1(xa2 + 2xaxi1)
(b) W2 = ∫-xi2-(xi2 + xa)k2xdx
= ½k2[(–xi2 + xa)2 – xi22]
W2 = ½ k2(xa2 – 2xaxi2)
(c) Before the horizontal force is applied, the springs exert equal forces:
k1xi1 = k2xi2
xi2 = k1xi1/k2
(d) the total work done by the force Fapp is
∑W = W1 + W2
= ½ k1(xa2 + 2xaxi1) + ½ k2(xa2 – 2xaxi2)
= ½ k1xa2 + xaxi1 + ½ k2xa2 –xa(k1xi1/k2)
∑W = ½ (k1 + k2)xa2
Problem#3
As the driver steps on the gas pedal, a car of mass 1 160 kg accelerates from rest. During the first few seconds of motion, the car’s acceleration increases with time according to the expression
a = (1.16 m/s3)t – (0.210 m/s4)t2 + (0.240 m/s5)t3
(a) What work is done by the wheels on the car during the interval from t = 0 to t = 2.50 s? (b) What is the output power of the wheels at the instant t = 2.50 s?
Answer:
(a) Work is done by the wheels on the car during the interval from t = 0 to t = 2.50 s, we use
∑W = ∆K
∑W = ½ mvf2 – 0
with
v = ∫adt = ∫{(1.16 m/s3)t – (0.210 m/s4)t2 + (0.240 m/s5)t3}dt
= (0.56t2 – 0.0700t3 + 0.0600t4)│0t
= 0.56t2 – 0.0700t3 + 0.0600t4
at t = 0, vi = 0. At t = 2.5 s,
vf = 0.56(2.5)2 – 0.0700(2.5)3 + 0.0600(2.5)4 = 4.88 m/s
then
∑W = ½ (1160 kg)(4.88 m/s)2 = 13.8 kJ
(b) the output power of the wheels at the instant t = 2.50 s, we use
P = F.v
Through the axles the wheels exert on the chassis force F = ma and
a = (1.16 m/s3)(2.5 s) – (0.210 m/s4)(2.5 s)2 + (0.240 m/s5)(2.5 s)3 = 5.34 m/s2
then F = (1160 kg)(5.34 m/s2) = 6.19 x 103 N
and inject power
P = (6.19 x 103 N)(4.88 m/s) = 30.2 kW
Problem#4
A particle is attached between two identical springs on a horizontal frictionless table. Both springs have spring constant k and are initially unstressed. (a) If the particle is pulled a distance x along a direction perpendicular to the initial configuration of the springs, as in Figure 2, show that the force exerted by the springs on the particle is
F = –2kx{1 – L/(x2 + L2)1/2}i
(b) Determine the amount of work done by this force in moving the particle from x = A to x = 0.
Answer:
(a) The new length of each spring is (x2 + L2)1/2, so its extension is (x2 + L2)1/2 – L and the force it exerts is
k{(x2 + L2)1/2 – L }
toward its fixed end. The y components of the two spring forces add to zero. Their x components add to
F = –2ik{(x2 + L2)1/2 – L}[x/(x2 + L2)1/2]
F = –2kx{1 – L/(x2 + L2)1/2}i
(b) the amount of work done by this force in moving the particle from x = A to x = 0 is
W = ∫Fxdx
= ∫A0[–2kx{1 – L/(x2 + L2)1/2}]dx
= ∫A0(–2kx)dx + ∫A0{kL/(x2 + L2)1/2}dx
= –kx2│A0 + 2kL(x2 + L2)1/2│A0
= 0 + kA2 + 2kL2 – 2kL(A2 + L2)1/2
W = kA2 + 2kL{L – (A2 + L2)1/2}
Problem#5
A rocket body of mass M will fall out of the sky with terminal speed vT after its fuel is used up. What power output must the rocket engine produce if the rocket is to fly (a) at its terminal speed straight up; (b) at three times the terminal speed straight down? In both cases assume that the mass of the fuel and oxidizer remaining in the rocket is negligible compared to M. Assume that the force of air resistance is proportional to the square of the rocket’s speed.
Answer:
For the rocket falling at terminal speed we have
∑F = ma
+R – Mg = 0
R = Mg
½ DρAvT2 = Mg
(a) For the rocket with engine exerting thrust T and flying up at the same speed,
∑F = Ma
T – Mg – R = 0
T = 2Mg
The engine power is
P = F.v = T.vT = 2MgvT
(b) For the rocket with engine exerting thrust Tb and flying down steadily at 3vT,
Rb = ½ DρA(3vT)2 = 9Mg
∑F = ma
–Tb – Mg + 9Mg = 0
Tb = 8Mg
The engine power is
P = F.v = T.v
P = 8Mg(3vT) = 24MgvT
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