A hammer of mass m falls from rest off of a roof and drops a height H onto your head.
a) Assuming that the tool is in actual contact with your head for a time △t before it stops and slides off, what is the algebraic expression for the average force it exerts on your skull while stopping?
To determine the algebraic expression for the average force exerted on your skull by the hammer while stopping, we can use Newton's second law of motion.
Newton's second law states that the force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a). In this case, the mass of the hammer is given as m.
From the problem, we know that the hammer falls from rest and stops, so the final velocity (vf) of the hammer is zero. Using the kinematic equation vf^2 = vi^2 + 2ad, where vi is the initial velocity, a is the acceleration, and d is the distance, we can solve for the acceleration.
Since the initial velocity is zero, the equation simplifies to vf^2 = 2ad. Rearranging the equation, we get a = vf^2 / (2d).
Now, let's find the time (Δt) it takes for the hammer to come to a stop. The equation to calculate the time is Δt = vf / a, where vf is the final velocity and a is the acceleration. Since vf is zero, the time is simply Δt = 0 / a = 0.
The average force (F_avg) exerted on your skull can be obtained by rearranging Newton's second law: F_avg = m * a.
Substituting the value of a obtained previously, we get F_avg = m * (0 / (2d)) = 0.
Therefore, the algebraic expression for the average force the hammer exerts on your skull while stopping is F_avg = 0.
mgH=mv²/2
v=sqrt(2gH)
F=Δp/Δt= Δ(mv)/ Δt=m Δ(v)/Δt=
=m[sqrt(2gH)-0]/Δt= m[sqrt(2gH)]/ Δt