a 60 kg student jumps from a table 1.2m high. When his feet hit the floor he bends his knees so that the stopping time is .06s. What is the average force that the floor exerts on his body in order to stop him?

To find the average force exerted by the floor on the student, we can use Newton's second law of motion, which states that force equals mass multiplied by acceleration (F = m * a).

1. First, we need to calculate the acceleration. We can use the equation of motion for vertical free fall: s = ut + 0.5 * a * t^2. In this equation, s represents the vertical distance fallen, u is the initial vertical velocity (which is 0 since the student is initially at rest on the table), t is the stopping time, and a is the acceleration. Rearranging the equation, we get: a = 2s / t^2.

Given that the student falls from a height of 1.2m and the stopping time is 0.06s, we can substitute these values into the equation: a = 2 * 1.2 / (0.06)^2.

2. Now that we have the acceleration, we can calculate the force exerted by the floor using Newton's second law. F = m * a, where m is the mass of the student, which is 60 kg: F = 60 * a.

By substituting the acceleration value into the equation, we get: F = 60 * (2 * 1.2 / (0.06)^2).

Now, let's calculate the force exerted by the floor on the student:

F = 60 * (2 * 1.2 / 0.06^2)
= 60 * (2 * 1.2 / 0.0036)
= 60 * (2 * 1.2 * (1 / 0.0036))
= 60 * (2.4 * 277.7778)
= 60 * 666.6667
≈ 40000 N

Therefore, the average force exerted by the floor on the student is approximately 40000 Newtons.

mgh=1/2 m v^2

velocity at impact= sqrt (2gh)

force*time= mass*velocity
solve for force