Question

A 150 g block connected to a light spring with a force constant of k

Answer 1

A)

maximum force exerted by the spring is

F = k x = ( 3 N / m) ( 5*10^ -2 m ) = 0.15 N

B)

If F = ( 0.44) ( 0.15 N ) = 0.066 N

x = F / k = ( 0.066 N ) / ( 3 N /m ) = 0.022 m = 2.2 cm

c)

According to conservation of energy

( 1/ 2) k x ^2 = ( 1/ 2) m v ^2

v ^2 = k /m x^2

v = sqrt ( 3 N / m / 0.15 kg ) ( 5*10^ -2 m )

= 0.223 m/s

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If v = 0.44 ( 0.223 m/s ) = 0.098 m/s

   ( 1/2) k x ^2 = ( 1/ 2) m v ^2

   x ^2 = ( m / k ) v ^2

   x = sqrt ( 0.15 kg / 3 N /m ) ( 0.098 m/s )

   =0.022 m

= 2.2 cm