Question

A stretched spring has an elastic potential energy of 35 J when it is stretched 0.54 m. What is the spring constant of the spring? Round your answer to two significant figures

Answer 1

Answer:

Let the force constant of the spring be k.

Elastic potential energy of a spring is given by E =

2

1

kx

2

. ....(i)

Given , elastic potential energy of stretched spring is E = 50 x

2

....(ii)

Comparing equations (i) and (ii) :

2

1

kx

2

= 50 x

2

We get k = 100

m

2

J

.

Explanation:

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Answer 2

Answer:

The spring constant of the spring is 240.05 N/m according to the information on the question above. This problem can be solved using the spring potential energy formula which stated as PE = 1/2 * k * x^2 where PE is the spring's potential energy, k is the constant of the spring, and x is the stretched displacement. (Calculation: 240.05=35*2 / (0.54^2)

Explanation:

240 is the correct Ans

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