ch-7

mg yb

y_a ∆r

F

F

I G U R E

7.1

The work done by an external agent on the system of the book and the Earth as the book is

lifted from ya to yb is equal to

mgyb mgya.

Potential Energy

Chapter 7

yb

y_a ∆r

F

I G U R E

7.2

The work done by the gravitational force on the book as the book falls from yb to ya is equal to mgyb mgya.

h y vf yi = h Ugi = mgh Ki = 0 y = 0 Ug = 0 yf = y Ugf = mgy Kf = mv12 f2

F

IGURE

7.4

(Example 7.1) A ball is dropped from rest at a height h above the ground. Initially, the total energy of the ball–Earth system is gravitational potential

energy, equal to mgh when h 0

is at the ground. When the ball is at elevation y, the total system energy is the sum of kinetic and potential energies.

Active Figure 7.3

(Quick Quiz 7.3) Three identical balls are thrown with the same initial speed from the top of a building.

1 3 2

(a) θ R Actor Sandbag (b) mactor mactorg T mbag mbagg (c) T yi

F

I G U R E

7.5

(Example 7.2) (a) An actor uses some clever staging to make his entrance. (b) Free-body diagram for the actor at the bottom of the circular path. (c) Free-body diagram for the sandbag when it is just lifted from the floor.

Active Figure 7.6

(a) An undeformed spring on a frictionless horizontal surface. (b) A block of mass m is pushed against the spring, compressing it through a distance x. (c) When the block is released from rest, the elastic potential energy stored in the system is transformed to kinetic energy of the block.

x = 0 x m x = 0 v (c) (b) (a) Us = kx12 2 K_i = 0 Kf = mv12 2 U_s = 0 m m

F

I G U R E

7.7

The work done against the force of friction depends on the path taken as the book is moved from

to ; hence, friction is a

nonconservative force. The work required is greater along the brown path than the blue path.

30.0° v_f d = 1.00 m vi = 0

0.500 m

F

I G U R E

7.8

(Example 7.3) A crate slides down a ramp under the influence of gravity. The potential energy of the crate–Earth system decreases, whereas the kinetic energy of the crate increases.

2.00 m

n

Fg= m g

F

I G U R E

7.9

(Example 7.4) If the slide is frictionless, the speed of the child at the bottom depends only on the height of the slide.

E = – mv1₂ A2 x = 0 (a) (b) (c) vC = 0 (d) x_max E = – mv1₂ B2 + – kx1₂ B2 E = – mv1₂ D2 = – mv1₂ A2 E = – kx1 _max 2 vA vB xB vD = –vA 2

F

I G U R E

7.10

(

Example 7.5) A block sliding on a smooth, horizontal surface collides with a light spring. (a) Initially, the mechanical energy of the system is all kinetic energy. (b) The mechanical energy is the sum of the kinetic energy of the block and the elastic potential energy in the spring. (c) The mechanical energy is entirely potential energy. (d) The mechanical energy is transformed back to the kinetic energy of the block. The total energy of the block–spring system remains constant throughout the motion.

Solar radiation through windows

Solar radiation on roof and walls Electrical

transmission

Energy enters or leaves home by heat through walls, roof, floor, and windows

Leaks in walls, windows, and doors allow matter transfer

Underground gas lines– matter transfer

F

I G U R E

7.11

Energy enters and leaves a home by several mechanisms. The home can be modeled as a nonisolated system in steady state.

F_g Fg m rf r_i M_E R_E

F

I G U R E

7.12

As a particle of mass m

moves from to above the Earth’s

surface, the potential energy of the

particle–Earth system, given by

Equation 7.19, changes because of the change in the particle–Earth separation distance r from rito rf.

Earth R_E O –GMEm U_g r R_E ME

F

IGURE

7.13

Graph of the gravitational potential energy Ug versus r for a particle above the Earth’s surface. The potential energy of the system goes to zero as r approaches infinity.

1 2 3 r₁₂ r₂₃ r₁₃

Active Figure 7.15

(a) Potential energy as a function of x for the block–spring system shown in part (b). The block oscillates between the turning points, which have the coordinates x xmax. The restoring force exerted by the spring always acts toward x 0, the position of stable equilibrium. E –x_max 0 Us x (a) x_max (b) m x = 0 = – kx1 2 2 U_s x_max Fs

0 x U Negative slope x > 0 Positive slope x < 0

F

I G U R E

7.16

A plot of U versus x for a particle that has a position of unstable equilibrium, located at x 0. For any finite displacement of the particle, the force on the particle is directed away

h R Figure _P7.5 20.0 m θ 60.0 m g P vi A B Figure P7.6 h 4.00 m m2 3.00 kg m₁ 5.00 kg Figure P7.9

(5.00, 5.00) m C B y x A O

Figure P7.14 Problems 7.14 and 7.15.

3.20 m m 2.00 m 5.00 m Figure P7.10 θ Figure P7.11

Figure P7.18 Problem 7.18 d θ k m Figure P7.16 (Gamma)

3.00 kg 5.00 kg Figure P7.23 3.00 m vi = 8.00 m/s 30.0° Figure P7.25 50.0 kg 100 kg 37.0° v Figure P7.28

4 U ( J) 6 2 0 –2 –4 2 4 6 8 x (m) Figure P7.39 0 +2 U ( J) +4 +6 +2 –2 –4 –6 2 4 6 r (mm) Figure P7.40 Top View L L x m k k x Figure P7.41