fisica1-youn-e-freedman-exercicios-resolvidos

7.65: a) The magnitude of the work done by friction is the kinetic energy of the
1 2
package at point B, or µ mgL = mv , or
k
(1 2)
µ =
gL
2
B
2
v B
k
=
2
(1 2)(4.80 m s)
=
2
(9.80 m s )(3.00 m)
0.392.
b)
W
other
= K B
−U
A
=
1
2
(0.200 kg)(4.80
= −0.832
J.
m s)
2
− (0.200
kg)(9.80
m s
2
)(1.60
m)
Equivalently, since K K = 0 , U + W + W = 0 , or
W
AB
A
=
B
A AB BC
= −U
−W
= mg( −(1.60
m) − (0.300)( −3.00
m)) = −0.832
A
BC
J.
1 2
7.66: Denote the distance the truck moves up the ramp by x. K
1
=
2
mv0
,
U
1
= mgLsinα , K = 2
0 , U mgxsin β
2
= and Wother
= −µ rmgx
cos β . From
W = K + U ) − ( K + U ) , and solving for x,
other
(
2 2 1 1
2
K1
+ mgL sin α ( v0
2g)
+ L sin α
x =
=
.
mg(sin
β + µ cos β)
sin β + µ cos β
r
r
7.67: a) Taking U ( 0) = 0 ,
x α 2 β 3
2
U ( x)
= ∫ Fx dx = x + x = (30.0 N m) x + (6.00
0 2 3
N
2
3
m ) x .
b)
and so
K
2
= U −U
1
2
2
= ((30.0 N m)(1.00 m)
2
− ((30.0 N m)(0.50 m)
= 27.75 J,
J)
v =
2(27.75 7.85 m s .
2 0.900 kg
=
3
+ (6.00 N m )(1.00 m) )
2
2
3
+ (6.00 N m )(0.50 m) )
7.68: The force increases both the gravitational potential energy of the block and the
potential energy of the spring. If the block is moved slowly, the kinetic energy can be
taken as constant, so the work done by the force is the increase in potential energy,
1 2
∆ U = mga sinθ
+ k(
aθ
) .
2