You use a slingshot to launch a 40-g stone upward at angle (not straight upward)

Question

You use a slingshot to launch a 40-g stone upward at angle (not straight upward). The spring has a spring constant of 100 N/m and an unstretched length of 10 cm. To launch the stone, you stretch the elastic to 30 cm. At the highest point in its flight, the stone moves at 5 m/s. Ignore friction and air resistance, and assume the initial height of 1.7 m. What is the height of the highest point of the flight? What is the initial stretch of the elastic? What is the initial potential energy of the stretched elastic? What is the initial kinetic energy of the stone before elastic is released? What is the gravitational e of the stone before the elastic is released? What is the total energy of the system just before the elastic is released? What is the total energy of the system once the stone is at the highest point? Why is the kinetic energy of the stone at the highest point of trajectory NOT zero? What is the kinetic energy of the stone at the highest point? What is the potential energy of the dart when it is at the highest point? What is the height of the dart at the highest point?

Explanation / Answer

I)

Initial stretch of the elastic is,

L = 30 cm - 10 cm

or, L = 20 cm

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II)

Initial potential energy of the stretched elastic is,

PE = (1/2)k(L)2 = (0.5)(100 N/m)(0.2 m)2

or, PE = 2 J

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III)

Before the elastic is released, the speed of the stone is zero. So the kinetic energy is zero.

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IV)

gravitational potential energy of the stone before the elastic is released is,

PE = mgh = (0.04kg)(9.8 m/s2)(1.7 m)

or, PE = 0.6664 J

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V)

Total energy of the system before the elastic is released = potential energy of the elastic + gravitational potential energy of the stone

So total energy is,

E = 2 J + 0.6664 J

or, E = 2.6664 J

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VI)

The system has no non-conservative force acting on it. So total energy will remain constant.

So total energy of the system when the stone is at highest point is,

E = 2.6664 J

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VII)

At the highest point, the stone moves at speed 5 m/s. Since the speed is not zero, kinetic energy is not zero.

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VIII)

The kinetic energy of the stone at the highest point is,

KE = (1/2)mv2 = (0.5)(0.04 kg)(5m/s)2

or, KE = 0.5 J

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IX)

At the highest point,

Total energy = kinetic energy + potential energy

or, 2.6664 J = 0.5 J + potential energy

So, potential energy = 2.1664 J

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X)

The potential energy of dart at highest point is PE = 2.1664 J

Let the height be h, then

mgh = 2.1664 J

or, (0.04 kg)(9.8 m/s2)h = 2.1664 J

or, h = 5.53 m

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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