A 0.40kg piece of putty is dropped from a height of 2.9m above a flat surface. When it hits the surface, the putty comes to rest in 0.32s .
What is the magnitude of the average force exerted on the putty by the surface?
I keep getting 9.4 as the answer, but it says it's wrong. I first calculated the velocity using sqrt(2gh) and got 7.539 m/s. And then solved for force using mv/t. Please tell me what I'm doing wrong.
To find the magnitude of the average force exerted on the putty by the surface, you need to use the impulse-momentum equation, which states that the change in momentum of an object is equal to the impulse applied to it. The impulse is the product of the force and the time it acts on the object.
Here's how you can calculate the average force:
1. First, calculate the initial velocity (vi) of the putty using the equation sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s²) and h is the height (2.9 m).
vi = sqrt(2 * 9.8 m/s² * 2.9 m)
vi ≈ 7.811 m/s (rounded to three decimal places)
2. Now, calculate the final velocity (vf) of the putty. Since it comes to rest, vf is 0 m/s.
vf = 0 m/s
3. Determine the change in momentum (Δp) by subtracting the initial momentum from the final momentum:
Δp = m(vf - vi)
Given the mass (m) of the putty is 0.40 kg:
Δp = 0.40 kg * (0 m/s - 7.811 m/s)
Δp ≈ -3.124 kg·m/s (rounded to three decimal places)
Note: The negative sign indicates a change in direction.
4. Finally, divide the change in momentum by the time (t) the force acts on the putty to find the average force (F):
F = Δp / t
Given the time (t) of 0.32 seconds:
F = (-3.124 kg·m/s) / 0.32 s
F ≈ -9.763 N (rounded to three decimal places)
The negative sign indicates that the force is acting in the opposite direction of the initial velocity.
Therefore, the magnitude of the average force exerted on the putty by the surface is approximately 9.763 N.
It seems like there may have been an error in your calculations. Double-check your numbers and ensure that you accounted for the negative sign in the final answer.