Chapter 08, Problem 013 Your answer is partially correct. Try again. A 7.2 o mar

Question

Chapter 08, Problem 013 Your answer is partially correct. Try again. A 7.2 o marble is fired vertically upward using a spring oun. The spring must be compressed 7.2 cm if the marble is to just reach a target 21 m above the marble's position on the compressed spring. (a) What is the change Au in the gravitational potential energy of the marble-Earth system during the 21 m ascent? b) what is the change us in the elastic potential energy of the spring during its launch of the marble? c what is the spring constant of the spring? Units (a) NumberT864 (b) Number .864 (c) Number 333.47 Click if you would like to Show Work for this question: Units7 Open Show Work

Explanation / Answer

x = 7.2 cm = 0.072 m

a) change in gravitational potential energy = m * g * h

change in gravitational potential energy = 0.0072 * 9.8 * 21

change in gravitational potential energy = 1.48 J

b)

When the marble is lost

change in elastic potential energy = - 1.48 J

c) let the spring constant is k

0.50 * k * 0.072^2 = 1.48

k = 572 N/m

the spring constant is 572 N/m

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