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S_A_V [24]
3 years ago
9

A 21 g ball is shot from a spring gun whose spring has a force constant of 579 N/m. The spring can be compressed 1 cm. How high

will the ball go if the gun is aimed vertically? The acceleration of gravity is 9.81 m/s 2 . Assume that at the end of the barrel, the spring’s displacement and the ball’s height are each zero
Physics
1 answer:
vodka [1.7K]3 years ago
7 0

Answer:

The answer  to the question is

The ball will go 0.14 meters high if the gun is aimed vertically

Explanation:

The energy in the spring → Energy, E = \frac{1}{2}kx^2

Where E = energy in the spring

k = Spring constant

x = Spring compression or stretch

Therefore E = \frac{1}{2} 579*0.01^2 = 0.02895 J

The spring energy is transferred  to the ball as kinetic energy based on the first law of thermodynamics which states that energy is neither created nor destroyed

Kinetic energy = KE = \frac{1}{2}mv^2

From which v = \sqrt{\frac{2*KE}{m} } = \sqrt{\frac{2*0.02895}{0.021} } = 1.66 m/s

from v² =u² - 2·a·S

Where v = final velocity = 0 m/s

u = initial velocity = 1.66 m/s

a = g = Acceleration due to gravity

S = height

Therefore 0 = 1.66² - 2×9.81×S

or S = 1.66² ÷ (2×9.81) = 0.14 m

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