fisica1-youn-e-freedman-exercicios-resolvidos

6.79: In terms of the bumper compression x and the initial speed v
0
, the necessary
relations are
1 2 1 2
kx = mv0
, kx < 5 mg.
2 2
Combining to eliminate k and then x, the two inequalties are
2
2
v
mg
x > and k < 25 .
2
5g
v
a) Using the given numbers,
2
(20.0 m /s)
x >
= 8.16 m,
2
5(9.80 m /s )
2 2
(1700 kg)(9.80 m / s )
4
k < 25
= 1.02×
10 N / m.
2
(20.0 m / s)
b) A distance of 8 m is not commonly available as space in which to stop a car.
6.80: The students do positive work, and the force that they exert makes an angle of
30 .0° with the direction of motion. Gravity does negative work, and is at an angle of
60 .0° with the chair’s motion, so the total work done is
2
W
tot
= ((600 N)cos30.0° − (85.0 kg)(9.80 m /s )cos60.0°)(2.50m)
= 257.8 J , and so the
speed at the top of the ramp is
2 2W
tot
2 2(257.8 J)
v
2
= v1
+ = (2.00 m /s) + = 3.17 m /s.
m
(85.0 kg)
Note that extra figures were kept in the intermediate calculation to avoid roundoff error.
6.81: a) At maximum compression, the spring (and hence the block) is not moving, so the
block has no kinetic energy. Therefore, the work done by the block is equal to its initial
1 2 1 2
kinetic energy, and the maximum compression is found from kX = mv , or
X
m
= v
k
b) Solving for v in terms of a known X,
v
k
= X
m
=
=
5.00 kg
(6.00 m s) = 0.600 m.
500 N m
500 N m
(0.150 m) = 1.50 m s.
5.00 kg
2
2