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A spring has a spring constant of 115 N/m. How much force is required to
stretch the spring 0.75 m past its natural length?
THERE
O A. 112 N
OB. 86.3 N
O C. 72.1 N
Ô D. 153 N
PAR
SUBMIT

Respuesta :

opudodennis

Answer:

Option B

86.3 N

Explanation:

From Hooke's law, F=kx where F is applied force, k is spring constant and x is the extension of the spring

Substituting 115 N/m for k and 0.75 for extension then

F= 115*0.75= 86.25 N

Approximately, 86.3 N

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