High School

An 9,100kg yacht was being carried by a 1.5m long high-quality spring. The spring stretches to 1.75m. Determine the spring constant if the friction is 14,000N opposite to the applied force.

Answer :

Final answer:

The spring constant of the high-quality spring is -56,000 N/m.

Explanation:

To determine the spring constant, 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
  • x is the displacement of the spring

In this case, we know the mass of the yacht is 9,100kg and the initial length of the spring is 1.5m. The final length of the spring is 1.75m.

First, we need to calculate the displacement of the spring:

x = final length - initial length

x = 1.75m - 1.5m

x = 0.25m

Next, we can use Hooke's Law to calculate the spring constant:

F = -kx

14,000N = -k * 0.25m

To isolate the spring constant, we divide both sides of the equation by -0.25m:

k = 14,000N / -0.25m

k = -56,000 N/m

Therefore, the spring constant is -56,000 N/m.

Learn more about calculating the spring constant here:

https://brainly.com/question/30763227

#SPJ14