A child of mass 40kg jumps off a wall and hits the ground at 4m/s he bends his knees and stops in 1s calculate the force required to slow him down how would this force be different if he didn't bend his knees and stopped in 0.1s
his acceleration is 4m/s^2
F = ma
1/10 the time means 10 times the acceleration.
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To calculate the force required to slow the child down, we can use Newton's second law of motion, which states that the force (F) acting on an object is equal to its mass (m) multiplied by its acceleration (a), or F = ma.
In this scenario, the child's initial velocity (u) is 4 m/s, his final velocity (v) is 0 m/s since he stops, and the time taken to stop (t) is 1 s. We can determine the acceleration using the formula:
a = (v - u) / t
Substituting the values, we have:
a = (0 - 4) / 1
a = -4 m/s²
Since the child is decelerating and the initial velocity is positive, the acceleration is negative.
Now, we can calculate the force required:
F = m * a
F = 40 kg * (-4 m/s²)
F = -160 N (since force is a vector quantity, a negative sign signifies that it acts in the opposite direction to the force applied)
Therefore, in order to slow down the child in 1 second, a force of 160 Newtons is required.
If the child didn't bend his knees and stopped in 0.1 second, we would need to calculate the force again. The process is the same, but the time is different.
Using the new time (t = 0.1 s), the acceleration (a) would be:
a = (0 - 4) / 0.1
a = -40 m/s²
Using this acceleration, the force required would be:
F = m * a
F = 40 kg * (-40 m/s²)
F = -1600 N
Therefore, if the child didn't bend his knees and stopped in 0.1 seconds, a force of 1600 Newtons would be required, which is ten times greater than the force required in the previous scenario.