Answer:
Therefore, the cat managed to get 625 meters before the leash pulled him back.
Explanation:
Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. In this case, the leash can be treated as a spring with an elastic constant (K) of 5 N/m.
The force exerted by the leash can be calculated using Hooke's Law: F = K * x, where F is the force, K is the elastic constant, and x is the displacement.
The work done by the force on the cat is equal to the force multiplied by the distance traveled: Work = F * d, where d is the distance traveled.
In this scenario, the force exerted by the leash is opposing the motion of the cat, so the work done by the force is negative. Thus, the equation becomes: Work = -F * d.
The work done by the force is equal to the change in kinetic energy of the cat, as per the work-energy principle.
So, -F * d = (1/2) * m * (v_final^2 - v_initial^2), where m is the mass of the cat, v_final is the final velocity (0 m/s in this case), and v_initial is the initial velocity (25 m/s in this case).
Plugging in the values, we have:
-5 * d = (1/2) * 10 * (0^2 - 25^2)
Simplifying the equation:
-5 * d = (1/2) * 10 * (-625)
-5 * d = -3125
Dividing both sides by -5, we get:
d = 625
