Question

A 100 kg deer runs through the forest away from a mountain lion. The deer changes speed from 3.00 m/s to 13.00 m/s in 4.00 seconds. What force does the deer exert on the ground as it accelerates?

Answer 1

The force exerted by a 100 kg deer as it accelerates from 3.00 m/s to 13.00 m/s in 4.00 seconds is calculated using Newton's second law of motion and is found to be 250 Newtons.

Step-by-step explanation:

The question is asking about the force exerted by a deer on the ground as it accelerates to escape a predator. We can calculate this force using Newton's second law of motion, F = ma, where F is the force, m is the mass of the deer, and a is the acceleration.

The acceleration can be found using the formula a = (v_f - v_i) / t, where v_f is the final velocity, v_i is the initial velocity, and t is the time taken to change velocity.

Given the mass of the deer is 100 kg, its initial velocity v_i is 3.00 m/s, its final velocity v_f is 13.00 m/s, and the time taken t is 4.00 seconds, we can find the acceleration as follows:

a = (13.00 m/s - 3.00 m/s) / 4.00 s = 2.50 m/s2.

With the acceleration known, we calculate the force:

F = m * a = 100 kg * 2.50 m/s2 = 250 N.

Therefore, the deer exerts a force of 250 Newtons on the ground as it accelerates.