Question

In: Physics

A shopper weighs 4.00 kg of apples on a supermarket scale whose spring obeys Hooke's law...

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 3.50 cm.

(a) What will the spring's extension be if 7.00 kg of oranges are weighed instead?
cm

(b) What is the total amount of work that the shopper must do to stretch this spring a total distance of 7.50 cm beyond its relaxed position?
J

Expert Solution

Mass of Apples ,m=4 kg

Stretching distance ,x=3.5 cm

So according to Hook's law,

Force, ...................(1)

K is spring constant.

so exert force of the spring is equal to weight of the mass.

F=mg........................(2)

From equation (1) and (2)

Part(a)

Mass of oranges,M=7 kg

If the spring extension is .

Then,

Answer : Spring extension is 6.125 cm whem moass of oranges is 7 kg.

Part (B)

Given Stretching length ,x=7.5 cm

So total amount of work is equal to potential energy in the spring.

Total work done is

Answer: the total amount of work is 3.15 J

genius_generous answered 2 years ago

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