A shopper weighs 4.00 kg of apples on a supermarket scale whose spring obeys Hooke's law and notes that the spring stretches a distance of 2.50 cm. (a) What will the spring's extension be if 6.50 kg of oranges are weighed instead? seenKey 4.06 Your response differs from the correct answer by more than 10%. Double check your calculations. cm (b) What is the total amount of work that the shopper must do to stretch this spring a total distance of 8.00 cm beyond its relaxed position?

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boffeemadrid boffeemadrid · Answer 1 · 2019-10-11T07:30:06+02:00

Answer:

a) 0.040625 m

b) 5.02272 J

Explanation:

k = Spring constant

x = Stretched length

F = Force

a)

[tex]F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{4\times 9.81}{0.025}\\\Rightarrow k=1569.6\ N/m[/tex]

[tex]F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{6.5\times 9.81}{1569.6}\\\Rightarrow x=0.040625\ m[/tex]

Extension of the spring would be 0.040625 m

b) Work done in a spring

[tex]W=\frac{1}{2}kx^2\\\Rightarrow W=\frac{1}{2}\times 1569.6\times 0.08^2\\\Rightarrow W=5.02272\ J[/tex]

The work done by the shopper to stretch this spring a total distance of 8.00 cm is 5.02272 J