fisica1-youn-e-freedman-exercicios-resolvidos

7.13: a) The force is applied parallel to the ramp, and hence parallel to the oven’s
motion, and so W = Fs = ( 110 N) (8.0 m) = 880 J. b) Because the applied force F r is
parallel to the ramp, the normal force is just that needed to balance the component of the
weight perpendicular to the ramp, n = wcos α,
and so the friction force is
f = µ mg cosα
and the work done by friction is
k
W
f
k
= −µ
mg cosα
s = −(0.25)
k
(10.0 kg) (9.80 m
2
s ) cos 37°
(8.0 m) = −157
J,
2
keeping an extra figure. c) mgs sin α = (10.0 kg)(9.80 m s )(8.0 m)sin 37°
= 472 J ,
again keeping an extra figure. d) 880 J − 472 J −157 J = 251J.
e) In the direction up the
ramp, the net force is
F − mg sin α − µ mg cosα
k
2
= 110 N − (10.0 kg)(9.80 m s )(sin 37° + (0.25) cos 37°
)
= 31.46 N,
2
so the acceleration is (31.46 N) 10.0 kg) = 3.15 m s . The speed after moving up the
2
ramp is v = 2as
= 2(3.15 m s ) (8.0 m) = 7.09 m s , and the kinetic energy is
(1 2) mv 2 = 252 J. (In the above, numerical results of specific parts may differ in the third
place if extra figures are not kept in the intermediate calculations.)
7.14: a) At the top of the swing, when the kinetic energy is zero, the potential energy
(with respect to the bottom of the circular arc) is mgl( 1− cos θ),
where l is the length of
the string and θ is the angle the string makes with the vertical. At the bottom of the
2
swing, this potential energy has become kinetic energy, so mgl (1 − cosθ)
= 1
mv , or
2
v = 2gl(1
− cosθ)
= 2(9.
80 m s ) (0.
80 m) (1 − cos 45°
) = 2.1m s . b) At 45 ° from the
vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the
2
radial component of the weight, or mg cos θ = (0.12 kg) (9.80 m s ) cos 45°
= 0.83 N. c)
At the bottom of the circle, the tension is the sum of the weight and the radial
acceleration,
2
mg + mv2 l = mg(1
+ 2(1 − cos 45°
)) = 1.86 N,
or 1.9 N to two figures. Note that this method does not use the intermediate calculation of
v.
2
7.15: Of the many ways to find energy in a spring in terms of the force and the
distance, one way (which avoids the intermediate calculation of the spring constant) is to
note that the energy is the product of the average force and the distance compressed or
extended. a) ( 1 2)(800 N)(0.200 m) = 80.0 J. b) The potential energy is proportional to
the square of the compression or extension; (80.0 J) (0.050 m 0.200 m) 5.0 J.
2 =