fisica1-youn-e-freedman-exercicios-resolvidos

7.51: K
1
+ U1
+ Wother
= K2
+ U
2
Point 1 is where he steps off the platform and point 2 is where he is stopped by the
1 2
cord. Let y = 0 at point 2. y1
= 41.0 m. Wother
= − kx , where x = 11.0 m is the amount
2
the cord is stretched at point 2.The cord does negative work.
1 2
K1 = K2
= U2
= 0,
so mgy
1
− kx = 0
2
and k = 631N/m.
Now apply F = kx to the test pulls:
F = kx so x = F k = 0.602 m.
7.52: For the skier to be moving at no more than 30 .0 m s ; his kinetic energy at the
bottom of the ramp can be no bigger than
2
2
mv (85.0 kg)(30.0 m s)
=
= 38,250 J
2
2
Friction does − 4000 J of work on him during his run, which means his combined PE
and KE at the top of the ramp must be no more than 38 ,250 J + 4000 J = 42,250 J. His KE
at the top is
2
2
mv (85.0 kg)(2.0 m s)
=
= 170 J
2
2
His PE at the top should thus be no more than 42 ,250 J − 170 J = 42,080 J,
which gives a
height above the bottom of the ramp of
42,080 J 42,080 J
h = =
= 50.5 m.
2
mg (85.0 kg)(9.80 m s )
7.53: The net work done during the trip down the barrel is the sum of the energy stored
in the spring, the (negative) work done by friction and the (negative) work done by
1 2 1 2
gravity. Using kx = ( F k)
, the performer’s kinetic energy at the top of the barrel is
2
2
K
2
1 (4400 N)
2
3
=
− (40 N)(4.0 m) − (60 kg)(9.80 m s )(2.5 m) = 7.17 × 10
2 1100 N m
J,
2(7.17×
10 J)
and his speed is = 15.5 m s.
60 kg
3