A ball with a mass of #240 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #12 (kg)/s^2# and was compressed by #1/2 m# when the ball was released. How high will the ball go?
The height is
The spring constant is The compression is The potential energy is This potential energy will be converted to kinetic energy when the spring is released The initial velocity is Resolving in the vertical direction We apply the equation of motion At the greatest height, and So,
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To find the maximum height ( h ) reached by the ball, we can use the principle of conservation of mechanical energy:
[ PE_{initial} + KE_{initial} = PE_{final} + KE_{final} ]
where ( PE ) is the potential energy and ( KE ) is the kinetic energy.
At the point of release, all the energy is potential energy stored in the spring:
[ PE_{initial} = \frac{1}{2} k x^2 ]
where ( k ) is the spring constant and ( x ) is the compression of the spring.
At maximum height, all the energy is potential energy:
[ PE_{final} = mgh ]
where ( m ) is the mass of the ball, ( g ) is the acceleration due to gravity, and ( h ) is the height reached.
Since the ball is projected vertically, the initial kinetic energy is zero.
[ KE_{initial} = 0 ]
At maximum height, the final kinetic energy is also zero.
[ KE_{final} = 0 ]
So, the equation becomes:
[ \frac{1}{2} k x^2 = mgh ]
Solve for ( h ):
[ h = \frac{1}{2} \times \frac{k x^2}{mg} ]
Given that ( m = 240 ) g, ( k = 12 ) (kg)/s(^2), ( x = \frac{1}{2} ) m, and ( g = 9.8 ) m/s(^2), we can substitute these values into the equation to find the height ( h ):
[ h = \frac{1}{2} \times \frac{12 \times (\frac{1}{2})^2}{0.240 \times 9.8} ]
[ h = \frac{1}{2} \times \frac{12 \times \frac{1}{4}}{2.352} ]
[ h = \frac{1}{2} \times \frac{3}{2.352} ]
[ h ≈ \frac{3}{4.704} ]
[ h ≈ 0.6375 , \text{m} ]
Therefore, the ball will reach a height of approximately 0.6375 meters.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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