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how far up the 15 degree incline of a pinball table will a 0.1 kg pinball move after it is launched? The spring constant is 100 N/m and is compressed by 0.08m.

User Tsang
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1 Answer

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We can solve this problem by using the principle of energy conservation. The mechanical energy of the system (kinetic + potential energy) will be conserved since we are neglecting friction and air resistance.

The initial potential energy of the spring when it is compressed is given by 1/2*k*x^2 where k is the spring constant and x is the distance it is compressed.

The final kinetic energy of the ball after the spring is released will be zero (assuming it comes to rest), but it will have potential energy due to being a certain height up the incline. The potential energy in this case is m*g*h, where m is the mass of the ball, g is the acceleration due to gravity (approximated as 9.8 m/s²), and h is the height.

Setting the potential energy equal to each other (spring potential energy = gravitational potential energy) and solving for h gives us:

1/2 * 100 N/m * (0.08 m)^2 = 0.1 kg * 9.8 m/s² * h

Solving this equation:

1/2 * 100 * 0.0064 = 0.1 * 9.8 * h

0.32 N*m = 0.98 * h

h = 0.32 N*m / 0.98 m/s² ≈ 0.33 m

The ball will move up to 0.33 m on the incline before coming to a stop.

Remember that this analysis assumes a frictionless, air resistance-less system, and neglects any energy that might be lost during the conversion of potential to kinetic energy. In real-world situations, these factors could significantly affect the results.

User Carstenbauer
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