Question

In 0.988 s, a 12.3-kg block is pulled through a distance of 4.85 m on a frictionless horizontal surface, starting from rest. the block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. the spring constant of the spring is 467 n/m. by how much does the spring stretch?

Answer 1

To calculate the amount by which the spring stretches, we can use Hooke's Law which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The spring stretches by 0.258 meters.

Step-by-step explanation:

To calculate the amount by which the spring stretches, we can use Hooke's Law which states that the force exerted by a spring is directly proportional to the displacement of the spring 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.

In this case, we are given the spring constant (k = 467 N/m) and the mass of the block (m = 12.3 kg).

The force exerted by the spring is equal to the weight of the block, which can be calculated using the formula F = mg, where g is the acceleration due to gravity (9.8 m/s^2).

By rearranging the formula for Hooke's law, we can solve for x:

x = F / k = (mg) / k = (12.3 kg * 9.8 m/s^2) / 467 N/m = 0.258 m

Therefore, the spring stretches by 0.258 meters.