MOMENTUM
Question:
A 20,000 N truck is acted upon by a force that decreases its speed from 10m/s to 5m/s in 5seconds. What is the magnitude of the force?
Solution:
First, we have to find the mass of the truck, we are given the weight which is m*g. so we can divide 20,000N / 9.8 = 2040.82 g (mass).
Mass*velocity= momentum. We need to find the cahnge in momentum divided by the time (s) to calculate the force.
Work:
(mVf-mVi) /t = F
m(Vf-Vi) /t =F
2040.82 [(5m/s-10m/s)] /5s =F
2040.82 (-1) = -2040.82
which is the magnitude of the force
Question 2)
A 6N force acts on a 3kg object fo 10seconds. What will be the final velocity of the object if its initial velocity was 10m/s?
First, we can find the acceleration using F=ma
6N=3kg*a
Using algebra: a=2m/s^2
Because acceleration multiplied by time = (delta)velocity,
a*t=v ---> 2m/s*10 =20m/s (change in velocity)
So, 20m/s +10m/s =30m/s
(delta v) (v initial)
OR
Vf=Vi+at
Vf=10m/s + 2m/s (10s)
Vf= 30m/s
A 20,000 N truck is acted upon by a force that decreases its speed from 10m/s to 5m/s in 5seconds. What is the magnitude of the force?
Solution:
First, we have to find the mass of the truck, we are given the weight which is m*g. so we can divide 20,000N / 9.8 = 2040.82 g (mass).
Mass*velocity= momentum. We need to find the cahnge in momentum divided by the time (s) to calculate the force.
Work:
(mVf-mVi) /t = F
m(Vf-Vi) /t =F
2040.82 [(5m/s-10m/s)] /5s =F
2040.82 (-1) = -2040.82
which is the magnitude of the force
Question 2)
A 6N force acts on a 3kg object fo 10seconds. What will be the final velocity of the object if its initial velocity was 10m/s?
First, we can find the acceleration using F=ma
6N=3kg*a
Using algebra: a=2m/s^2
Because acceleration multiplied by time = (delta)velocity,
a*t=v ---> 2m/s*10 =20m/s (change in velocity)
So, 20m/s +10m/s =30m/s
(delta v) (v initial)
OR
Vf=Vi+at
Vf=10m/s + 2m/s (10s)
Vf= 30m/s