Question

A 10.0-kg mass hangs from a spring that has the spring constant 535 N/m. Find the position of the end of the spring away from its rest position. (Use g=9.80 m/s^2.)

a) 0.35 m
b) 0.50 m
c) 0.70 m
d) 0.85 m

Answer 1

The position of the end of the spring away from its rest position is approximately 0.183 m.

Step-by-step explanation:

To find the position of the end of the spring away from its rest position, we can use Hooke's Law which states that the force exerted by a spring is proportional to the displacement from its equilibrium position.

The formula for Hooke's Law is F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position. Rearranging the formula, we get x = -F / k.

In this case, the mass hanging from the spring experiences a gravitational force of mg = 10.0 kg * 9.80 m/s^2 = 98 N. Therefore, the position of the end of the spring away from its rest position is x = -98 N / 535 N/m = -0.183 m.

Since the displacement is negative, the correct answer is the position in the opposite direction, which is approximately 0.183 m. Therefore, none of the given options (a) 0.35 m, (b) 0.50 m, (c) 0.70 m, or (d) 0.85 m, are correct.