change in momentum = force * time
3 v = 20*.25
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?
2 answers
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)
==================
check
3 v = 20 * t
3 sqrt (10/3)= 20 * .5/sqrt(10/3)
3 (10/3) = 20 * .5 yes
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)
==================
check
3 v = 20 * t
3 sqrt (10/3)= 20 * .5/sqrt(10/3)
3 (10/3) = 20 * .5 yes
2 answers