In Figure 10-1, if the force exerted on a 3.0-kg backpack that is initally at rest is 20.0 N and the distance it acts over is 0.25 m, what is the final speed of the backpack?
change in momentum = force * time
3 v = 20*.25
I multiplied by distance not time
average speed = v/2
distance = .25 = t (v/2)
t = .5/v
then
3 v = 20 *.5/v
3 v^2 =10
v^2 = 10/3
v = sqrt(10/3)
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check
3 v = 20 * t
3 sqrt (10/3)= 20 * .5/sqrt(10/3)
3 (10/3) = 20 * .5 yes
To solve this problem, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. The formula for work is:
work = force × distance × cos(θ)
We can assume that the force exerted on the backpack is in the same direction as the displacement, so we can simplify the formula to:
work = force × distance
To find the work done on the backpack, we can substitute the given values into the formula:
work = 20.0 N × 0.25 m = 5.0 J
Since the backpack is initially at rest, its initial kinetic energy is zero. Therefore, the work done on the backpack is equal to its change in kinetic energy:
work = ΔKE
ΔKE = 5.0 J
The formula for kinetic energy is:
KE = 0.5 × mass × velocity^2
We can rearrange this equation to solve for the final velocity:
velocity = √(2 × KE / mass)
Substituting the values into the formula:
velocity = √(2 × 5.0 J / 3.0 kg)
= √(10 / 3) m/s
≈ 1.83 m/s
Therefore, the final speed of the backpack is approximately 1.83 m/s.