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Try these practice problems. If you need to, click on the hint button. When you're done, click on the check your answer link to see the final answer.

Question 1

A spring-loaded dart gun contains a spring with k = 10 N m . When fully inserted, the 0.002 kg dart compresses the spring 0.05 m. How much energy is stored in the spring?

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Identify known variables: k = 10 N m ; x = 0.05 m
Equation: PE_s = 1 2 kx²

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PE_s = 1 2 kx²

= 1 2 (10)(.05)²

= 1.25 × 10^-2J

Question 2

After the dart is released, all the energy becomes kinetic energy. What is the speed of the dart after it is released? Friction and air resistance are negligible.

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Identify Known variables: KE = 1.25 * 10^-2 J, m = 0.002 kg
Equation: KE = 1 2 mv²

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KE = 1 2 mv²

2ke m = v²

$Square root of entire fraction - numerator:2KE, denominator: m$

$Square root of entire fraction - numerator: 2(1.25 * 10 ^ -2) denominator: .002$

v = 3.5355 m/s

Question 3

Assume the same gun is fired straight up. If all of the energy becomes gravitational potential energy, how high will the dart go?

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Identify Known variables: PE_g = 1.25 * 10^-2J, m = 0.002 kg
Equation: PE_g = mgh

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PE_g = mgh

PE_g mg = h

h = 1.25 × 10^-2 (.002)(9.8)

h = 0.64 m

Student on a skateboard at various locations on a ramp. Data given in image: mass of student is 50 kilograms; at point 1, velocity is 25 m/s and h = 0; at point 2 velocity is 24 m/s and h = 2.5 meters; at point 3, velocity is 0 and h is unknown.

Adapted from: image 051, Energy files, Mr. Fizix

Question 4

Use the illustration provided to answer the next three problems. The height is 0 m at point 1. Friction is negligible.

Determine the potential energy PE, the kinetic energy KE, and the total energy ME at point 1.

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Identify known variables: m = 50 kg; v = 25 m/s; h = 0 m
Equations: PE_g = mgh; KE = 1 2 mv²; ME = KE + PE

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PE_g = mgh

PE_g = (50)(9.8)(0) = 0J

KE = 1 2 mv²

KE = 1 2 (50)(25)²

KE = 15625 J

ME = KE + PE

ME = 15625 + 0 = 15625 J

Question 5

Determine the potential energy PE, the kinetic energy KE, the mechanical energy ME at point 2.

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Identify known variables: m = 50 kg; v = 24 m/s; h = 2.5 m
Equations: PE_g = mgh; KE = 1 2 mv²; ME = KE + PE

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PE_g = mgh

PE_g = (50)(9.8)(2.5) = 1225 J

KE = 1 2 mv²

KE = 1 2 (50)(24)²

KE = 14400 J

ME = KE + PE

ME = 14400 + 1225 = 15625 J

(ENERGY STAYED THE SAME!!)

Question 6

Determine the potential energy PE, the kinetic energy KE, and the height h at point 3.

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ME of the skater on the skateboard remains constant with no friction, therefore, ME = 15625 J

Identify known variables: m = 50 kg; ME = 15625 J; v = 0 m/s
Equations: PE_g = mgh; KE = 1 2 mv²; ME = KE + PE

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Check Your Answer

With v = 0 m/s, KE = 0 J

ME = PE + KE

15625 = PE + 0

PE_g = 15625

PE_g = mgh

15625 = (50) (9.8) h

H = 15625/490 = 31.9 m