A force of 547 N keeps a certain spring stretched a distance of 0.300 m. A- what

Question

A force of 547 N keeps a certain spring stretched a distance of 0.300 m.
A- what is the potential energy of the spring when it is stretched 0.300 m?
B- what is the potential energy when it is compressed 6.00 cm? A force of 547 N keeps a certain spring stretched a distance of 0.300 m.
A- what is the potential energy of the spring when it is stretched 0.300 m?
B- what is the potential energy when it is compressed 6.00 cm?
A- what is the potential energy of the spring when it is stretched 0.300 m?
B- what is the potential energy when it is compressed 6.00 cm?

Explanation / Answer

here,

force required , f = 547 N

distance , x = 0.3 m

let the spring constant be k

f = k*x

547 = k * 0.3

k = 1823.33 N/m

A-

the potential energy of the spring when it is stretched 0.300 m , PE = 0.5 * k*x^2

PE = 0.5 * 1823.33 * 0.3^2

PE = 82.05 J

the potential energy of the spring when it is stretched 0.300 m is 82.05 J

B-

the potential energy when it is compressed 6.00 cm , PE =0.5 * k*x^2

PE = 0.5 * 1823.33 * 0.06^2

PE = 3.28 J

the potential energy when it is compressed 6.00 cm is 3.28 J

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