Problem#1
A 4.00-kg particle is subject to a total force that varies with position as shown in Figure 1. The particle starts from rest at x = 0. What is its speed at (a) x = 5.00 m, (b) x = 10.0 m, (c) x = 15.0 m?
Answer:
(a) We use
∆K = Kf – Ki = ½ mvf2 – 0 = ∑W = (area under curve from x = 0 to x = 5.00 m)
area = ½ x 5 x 3 J = 7.50 J, then
vf2 = 2∑W/m = 2(7.50 J)/(4.00 kg)
vf = 1.94 m/s
(b) ∆K = Kf – Ki = ½ mvf2 – 0 = ∑W = (area under curve from x = 0 to x = 10.0 m)
area = ½ x (5 + 10) x 3 J = 22.5 J, then
vf2 = 2∑W/m = 2(22.5 J)/(4.00 kg)
vf = 3.35 m/s
(c) ∆K = Kf – Ki = ½ mvf2 – 0 = ∑W = (area under curve from x = 0 to x = 15.0 m)
area = ½ x (5 + 15) x 3 J = 30.0 J, then
vf2 = 2∑W/m = 2(30.0 J)/(4.00 kg)
vf = 3.87 m/s
Problem#2
You can think of the work–kinetic energy theorem as a second theory of motion, parallel to Newton’s laws in describing how outside influences affect the motion of an object. In this problem, solve parts (a) and (b) separately from parts (c) and (d) to compare the predictions of the two theories. In a rifle barrel, a 15.0-g bullet is accelerated from rest to a speed of 780 m/s. (a) Find the work that is done on the bullet. (b) If the rifle barrel is 72.0 cm long, find the magnitude of the average total force that acted on it, as F = W/(∆r cos θ). (c) Find the constant acceleration of a bullet that starts from rest and gains a speed of 780 m/s over a distance of 72.0 cm. (d) If the bullet has mass 15.0 g, find the total force that acted on it as ∑F = ma.
Answer:
(a) the work that is done on the bullet, we use
Ki + ∑W = Kf = ½ mvf2
0 + ∑W = ½ (15.0 x 10-3 kg)(780 m/s)2 = 4.56 kJ
(b) the magnitude of the average total force that acted on it, as
F = W/(∆r cos θ)
F = (4.56 x 103 J)/(0.720 m cos00) = 6.34 x 103 N
(c) the constant acceleration of a bullet that starts from rest and gains a speed of 780 m/s over a distance of 72.0 cm is
a = (vf2 – vi2)/2xf
a = [(780 m/s)2 – 0]/[2(0.720 m)] = 422 m/s2
(d) the total force that acted on it as
∑F = ma = (15.0 x 10-3 kg)(422 m/s2) = 6.34 x 103 N
Problem#3
In the neck of the picture tube of a certain black-and-white television set, an electron gun contains two charged metallic plates 2.80 cm apart. An electric force accelerates each electron in the beam from rest to 9.60% of the speed of light over this distance. (a) Determine the kinetic energy of the
electron as it leaves the electron gun. Electrons carry this energy to a phosphorescent material on the inner surface of the television screen, making it glow. For an electron passing between the plates in the electron gun, determine (b) the magnitude of the constant electric force acting on the electron, (c) the acceleration, and (d) the time of flight.
Answer:
(a) the kinetic energy of the electron as it leaves the electron gun is
vf = 9.60%(3 x 108 m/s) = 2.88 x 107 m/s, the
Kf = ½ mvf2 = ½ (9.11 x 10-31 kg)(2.88 x 107 m/s)2 = 3.78 x 10-16 J
(b) the constant electric force acting on the electron, given by
Ki + W = Kf
0 + F∆r cosθ = Kf
F(0.028 m) cos00 = 3.78 x 10-16 J
F = 1.35 x 10-14 N
(c) the acceleration is
a = ∑F/m = (1.35 x 10-14 N)/(9.11 x 10-31 kg) = 1.48 x 1016 m/s2
(d) the time of flight is
vxf = vxi + at
2.88 x 107 m/s = 0 + (1.48 x 1016 m/s2)t
t = 1.94 x 10-9 s = 1.94 ns
Post a Comment for "Kinetic Energy and the Work–Kinetic Energy Theorem Problems and Solutions 2"