When a hammer strikes a nail, it is compressed along its length. Suppose that a steel nail 9.00 cm long and 3.00 mm in diameter is compressed an average of 65.0 µm by a 0.490 kg hammer during a 1.20 ms collision.
a) What is the average force exerted by the hammer on the nail?
I have tried to use F=ma and solve for a using dx/(dt)^2 (or dv/dt). When I entered 6.5e-5/(1.2e-3)^2 for acceleration and then multiplied the acceleration by .490kg, the answer was incorrect.
To find the average force exerted by the hammer on the nail, you can use the equation F = Δp/Δt, where F is the force, Δp is the change in momentum, and Δt is the change in time.
First, let's calculate the change in momentum:
Δp = m * Δv
The change in velocity can be obtained by dividing the change in length of the nail by the collision time:
Δv = Δx/Δt
Here, Δx is the change in length of the nail, which is given as 65.0 µm (or 6.5e-5 m), and Δt is the collision time, which is given as 1.20 ms (or 1.20e-3 s).
Plugging in the values:
Δv = 6.5e-5 m / 1.20e-3 s
Now, substitute the change in velocity into the equation for change in momentum:
Δp = m * (6.5e-5 m / 1.20e-3 s)
Next, we need to find the mass of the nail, which is given as 0.490 kg.
Finally, substitute the calculated change in momentum and the given collision time into the equation for average force:
F = Δp/Δt = (m * (6.5e-5 m / 1.20e-3 s)) / 1.20e-3 s = m * 6.5e-5 m / (1.20e-3 s)^2
Plugging in the values:
F = 0.490 kg * 6.5e-5 m / (1.20e-3 s)^2
Now, calculate the result using a calculator:
F ≈ 22.64 N
Therefore, the average force exerted by the hammer on the nail is approximately 22.64 Newtons.