fisica1-youn-e-freedman-exercicios-resolvidos

2
7.16: U = 1
ky , where y is the vertical distance the spring is stretched when the weight
2
mg
F
w = mg is suspended. y = ,
k
and k = ,
x
where x and F are the quantities that
“calibrate” the spring. Combining,
2
2 2
1 ( mg)
1 ((60.0 kg) (9.80 m s ))
U = =
= 36.0 J
2 F x 2 (720 N 0.150 m)
2 U
7.17: a) Solving Eq. (7.9) for x x =
2(3.20 J)
= 0.063 m.
, =
k (1600 N m)
b) Denote the initial height of the book as h and the maximum compression of the
spring by x. The final and initial kinetic energies are zero, and the book is initially a
height x + h above the point where the spring is maximally compressed. Equating initial
and final potential energies, 1 2
kx = mg(
x + h ).
2
This is a quadratic in x, the solution to
which is
mg ⎡ 2kh
⎤
x = ⎢1
± 1+
⎥
k ⎣ mg ⎦
2
(1.20 kg)(9.80 m s ) ⎡ 2(1600 N m) (0.80 m) ⎤
=
⎢1
± 1+
2 ⎥
(1600 N m) ⎣ (1.20 kg) (9.80 m s ) ⎦
= 0.116 m, − 0.101m.
The second (negative) root is not unphysical, but represents an extension rather than a
compression of the spring. To two figures, the compression is 0.12 m.
7.18: a) In going from rest in the slingshot’s pocket to rest at the maximum height, the
potential energy stored in the rubber band is converted to gravitational potential energy;
−3
2
U = mgy = (10×
10 kg)(9.80 m s ) (22.0 m) = 2.16 J.
b) Because gravitational potential energy is proportional to mass, the larger pebble
rises only 8.8 m.
c) The lack of air resistance and no deformation of the rubber band are two possible
assumptions.
7.19: The initial kinetic energy and the kinetic energy of the brick at its greatest height
1 2
are both zero. Equating initial and final potential energies, kx = mgh,
where
2
h is the
greatest height. Solving for h,
2
2
kx (1800 N m) (0.15 m)
h = =
= 1.7 m.
2
2 mg 2(1.20 kg) (9.80 m s )