A 0.37 kg piece of putty is dropped from a height of 3.0 m above a flat surface. When it hits the surface, the putty comes to rest in 0.25 s. What is the average force exerted on the putty by the surface?
magnitude N
direction
Impulse = Force * time = momentum change
Use the height to determine the velocity at impact, which you need for the momentum cha nge
.74
To find the average force exerted on the putty by the surface, you can use Newton's second law of motion, which states that the force acting on an object equals the mass of the object multiplied by its acceleration:
F = m * a
First, let's calculate the acceleration of the putty. We can use the equation for average acceleration:
a = (vf - vi) / t
Where:
vf = final velocity (0 m/s since the putty comes to rest)
vi = initial velocity
t = time taken to come to rest (0.25 s)
Next, we need to find the initial velocity. To do this, we can use the equation for free fall:
vf^2 = vi^2 + 2 * g * h
Where:
vf = final velocity (0 m/s)
vi = initial velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height (3.0 m)
Now we can solve for the initial velocity:
0^2 = vi^2 + 2 * 9.8 * 3.0
vi^2 = -2 * 9.8 * 3.0
vi^2 = -58.8
Since the initial velocity cannot be negative, we can ignore the negative sign:
vi = √(58.8) ≈ 7.671 m/s
Now we can calculate the acceleration:
a = (0 - 7.671) / 0.25
a ≈ -30.684 m/s^2
The negative sign indicates that the putty is decelerating.
Finally, we can calculate the average force:
F = m * a
F = 0.37 kg * -30.684 m/s^2
F ≈ -11.347 N
Since force is a vector quantity, the direction of the force is opposite to the direction of motion, which in this case is downwards (negative). Therefore, the magnitude of the average force exerted on the putty by the surface is approximately 11.347 N and the direction is downwards.