Ball and Spring mechanics

Classical Mechanics Level 2

A ball with a mass ( m m ) of 30 kg 30\text{ kg} , is placed on top of a frictionless hill at a height ( h h ) of 5 m 5\text{ m} . How much will the spring at the bottom of the hill be displaced in order to stop the ball if the spring constant ( k k ) is 30 N m 30 \frac{N}{m} ?

Note : A s s u m e g = 10 m s 2 U s p r i n g = 1 2 k x 2 U p o t e n t i a l = m g h Assume \space g = 10 \frac{m}{s^2} \\ U_{spring} = \frac{1}{2} k x^2 \\ U_{potential} = mgh

20 m 20\text{ m} 5 m 5\text{ m} 15 m 15\text{ m} 10 m 10\text{ m}

David Hontz by David Hontz , 26, USA

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1 solution

David Hontz
Jun 9, 2016

All the potential energy of the ball at the top of the hill will be transferred to the spring; therefore, U p o t e n t i a l = U s p r i n g m g h = 1 2 k x 2 x = 2 m g h k = 2 ( 30 k g ) ( 10 m s 2 ) ( 5 m ) 30 N m = 100 = 10 m U_{potential} = U_{spring} \\ mgh = \frac{1}{2} k x^2 \\ x = \sqrt{\frac{2mgh}{k}} = \sqrt{\frac{2(30kg)(10 \frac{m}{s^2})(5m)}{30\frac{N}{m}}} = \sqrt{100} = \boxed{10m}

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