Problem 7.42 A slingshot obeying Hooke\'s law is used to launch pebbles vertical

Question

Problem 7.42 A slingshot obeying Hooke's law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 18.2 cm against the elastic band, the pebble goes 5.0 m high. Part B How far must you pull it back so it will reach 10.0 m? Express your answer to three significant figures and include the appropriate units. Part C If you pull a pebble that is twice as heavy back 18.2 cm, how high will it go? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

given,

h=5m

x=18.2cm

now,

by using conservation of enrergy

1/2*k*x^2=mgh

====>

k/mg=2h/x^2 ------(1)

=2*5/0.182^2

k/mg=301.89

B)

here,

h=10 m

from eqn no (1)

k/mg=2h/x^2

301.89=2*10/x^2

x^2=20/301.89

====>

x=0.2574 m

=25.74 cm

C)

now,

m'=2m

x=18.2 cm

use,

k/m'g=2h'/x^2

k/2mg=2h'/x^2

k/mg=4h'/x^2

h'=(x^2/4)*(k/mg)

=(0.187^2/4)*(301.89)

=2.64 m

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