A 0.5 Kg pinball is initially at rest against a 120 N/m spring. The shooter is pulled back and has the spring compressed a distance of 0.2 m. The spring is released and the ball is shot up the ramp. It hits nothing and eventually comes to rest before it begins to roll down. We can ignore friction. The game board ramp is at an angle of 30o . How far did the ball travel on the board from the place of maximal compression to the first stop

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xero099 xero099 · Answer 1 · 2020-05-06T05:19:08+02:00

Answer:

[tex]\Delta s = 0.978\,m[/tex]

Explanation:

The pinball-spring system is modelled after the Principle of Energy Conservation:

[tex]U_{g,1} + U_{k,1} + K_{1} = U_{g,2} + U_{k,2} + K_{2}[/tex]

[tex]-(0.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right) \cdot \Delta h + \frac{1}{2}\cdot \left(120\,\frac{N}{m}\right)\cdot (-0.2\,m)^{2} = 0[/tex]

The height reached by the pinball is:

[tex]\Delta h = 0.489\,m[/tex]

The distance travelled by the pinball is:

[tex]\Delta s =\frac{0.489\,m}{\sin 30^{\circ}}[/tex]

[tex]\Delta s = 0.978\,m[/tex]