- #1
teclo
- 117
- 0
hi, I'm wondering if anyone could explain to me why one uses the following technique to solve this problem.
Let's say we've a 10 kg object. The force exerted on it is
F = 10 + 2t
The force acts on it for two (2) seconds. If it starts at rest, what is the final velocity?
I used this technique to solve the problem, after first making a mistake
dp = F(net) * dt
breaking that down to
dp = (10 + 2t) * dt
integration giving
10t + t^2 from 0 to 2
giving
24 kg * m/s
so
dp = 24 kg * m/s
m(v(2)-v(1)) = 24 kg * m/s
10 kg (v(2)-0) = 24 kg * m/s
v(2) = 2.4 m/s
this is the correct answer, however, initially i tried
m(v(2)-v(1)) = (10 + 2t) * dt
10 kg (v(2)-0) = 14N * 2s
resulting in a velocity of 2.8 m/s
why do i use integration before involving momentum is my question. thanks for anyone who might be able to help!
Let's say we've a 10 kg object. The force exerted on it is
F = 10 + 2t
The force acts on it for two (2) seconds. If it starts at rest, what is the final velocity?
I used this technique to solve the problem, after first making a mistake
dp = F(net) * dt
breaking that down to
dp = (10 + 2t) * dt
integration giving
10t + t^2 from 0 to 2
giving
24 kg * m/s
so
dp = 24 kg * m/s
m(v(2)-v(1)) = 24 kg * m/s
10 kg (v(2)-0) = 24 kg * m/s
v(2) = 2.4 m/s
this is the correct answer, however, initially i tried
m(v(2)-v(1)) = (10 + 2t) * dt
10 kg (v(2)-0) = 14N * 2s
resulting in a velocity of 2.8 m/s
why do i use integration before involving momentum is my question. thanks for anyone who might be able to help!