A plastic spring with spring constant 850 N/m has a relaxed length of 0.100 m. T

Question

A plastic spring with spring constant 850 N/m has a relaxed length of 0.100 m. The spring is positioned vertically on a table, and a charged plastic 1.00-kg sphere is placed on the top end of the spring. Another charged object is suspended above the sphere without making contact.

If the length of the spring is now 0.0950 m, what are the magnitude and direction of the electric force exerted on the sphere?

Express your answer with the appropriate units. Enter positive value if the force is directed upward and negative value if the force is directed downward.

Explanation / Answer

change in length of spring x = 0.1 - 0.0950 m = 0.005 m

spring force Fs= kx = 850 x 0.005 = 4.25 N

weight of body = mg = 1 x 9.8 = -9.8 N

for equillibrum electriic force Fe + weight + spring force Fs =0

Fe -9.8 + 4.25 = 0

Fe = 5.55N

the positive sign indicates its upwards

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