fisica1-youn-e-freedman-exercicios-resolvidos

6.14: (a)
(b)
W = ∆KE
− mgh =
v
0
=
1
2
mv
v
2
f
2
f
−
1
2
+ 2gh
mv
= (25.0 m / s)
= 30.3 m / s
W = ∆KE
2
0
2
+ 2(9.80 m / s
1 2 1 2
− mgh = mvf
− mv0
2 2
2 2
2
v0
− vf
(30.3 m /s) − 0
h = =
2
2g 2(9.80 ms/s )
= 46.8 m
2
)(15.0 m)
2
6.15: a) parallel to incline: force component = mg sinα
, down incline; displacement
= h/ sinα
, down incline
W = ( mg sinα )( h / sin ) = mgh perpendicular to incline: no displacement in this
||
α
direction, so W
⊥
= 0 .
W mg
= W|| + W⊥
= mgh , same as falling height h.
1 2
b) Wtot = K2 − K1
gives mgh =
2
mv and v = 2gh
, same as if had been dropped
from height h. The work done by gravity depends only on the vertical displacement of the
object.
When the slope angle is small, there is a small force component in the direction of
the displacement but a large displacement in this direction. When the slope angle is large,
the force component in the direction of the displacement along the incline is larger but
the displacement in this direction is smaller.
c) h = 15.0 m , so v = 2 gh = 17.1s
6.16: Doubling the speed increases the kinetic energy, and hence the magnitude of the
work done by friction, by a factor of four. With the stopping force given as being
independent of speed, the distance must also increase by a factor of four.
6.17: Barring a balk, the initial kinetic energy of the ball is zero, and so
2
2
W = (1/ 2) mv = (1/ 2)(0.145 kg)(32.0 m/s) = 74.2 J.