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6.101: a) Denote the position of a piece of the spring by l; l = 0 is the fixed point and<br />

l = L is the moving end of the spring. Then the velocity of the point corresponding to l,<br />

1<br />

denoted u, is u( l)<br />

= v (when the spring is moving, l will be a function of time, and so u<br />

L<br />

is an implicit function of time). The mass of a piece of length dl is dm = dl,<br />

and so<br />

and<br />

2 1 Mv<br />

dK = dmu =<br />

3<br />

2 2 L<br />

K =<br />

2<br />

1 2<br />

Mv<br />

2L<br />

2 L<br />

2<br />

∫ dK = ∫ l dl =<br />

3<br />

0<br />

l dl,<br />

1 2 1 2<br />

−2<br />

b) kx = mv ,<br />

2 2<br />

so v = ( k m)<br />

x = (3200 N m) (0.053 kg)(2.50 × 10 m) = 6.1m s. .<br />

c) With the mass of the spring included, the work that the spring does goes into the<br />

kinetic energies of both the ball and the spring, so 1 2 1 2 1 2<br />

kx = mv + Mv . Solving for v,<br />

v =<br />

d) Algebraically,<br />

k<br />

m + M<br />

x =<br />

3<br />

(3200 N m)<br />

(0.053 kg) + (0.243 kg)<br />

2<br />

2<br />

Mv<br />

6<br />

2<br />

.<br />

6<br />

(2.50 × 10<br />

3<br />

2<br />

1 2 (1 2) kx<br />

mv =<br />

= 0.40 J and<br />

2 (1 + M 3m)<br />

1 2<br />

Mv<br />

6<br />

2<br />

(1 2) kx<br />

=<br />

= 0.60 J.<br />

(1 + 3m<br />

M )<br />

−2<br />

L<br />

M<br />

m) = 3.9 m s.

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