Question

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0 cm. The body and spring are placed on a horizontal frictionless surface and rotated about the held end of the spring at 2.0 rev/s. How far is the spring stretched?

a) 2.47 cm
b) 3.82 cm
c) 5.26 cm
d) 6.10 cm

Answer 1

The spring is stretched by a distance of 0.01 m or 1.0 cm.

Step-by-step explanation:

To find the stretch of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from equilibrium. The force exerted by the spring can be calculated using the equation:

F = k * x

Where F is the force, k is the spring constant, and x is the displacement from equilibrium. In this case, the spring is extended 5.0 cm from its unstretched length of 4.0 cm, so the displacement is 5.0 cm - 4.0 cm = 1.0 cm = 0.01 m.

Plugging these values into the equation, we have:

F = (150 N/m) * (0.01 m) = 1.5 N

The force exerted by the spring is 1.5 N, which is equal to the weight of the body. Therefore, the spring is stretched by a distance of 0.01 m or 1.0 cm.