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pirateboy
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I'm a bit confused by this question. If you could help me check my work, that'd be great.
Thanks.
A 40kg object with an initial velocity \vec{v}_0 = (20 \text{m}/\text{s})\hat{i} is accelerated by a varying net force \vec{F}(t)=\left[ \left( 3.0 \text{N}/\text{s}^2\right)t^2 - \left( 4.0 \text{N}/\text{s}^2 \right)t + 10.0\text{N} \right]\hat{i} over a time interval of 3.0 seconds.
a) What is the net implulse (\vec{J}) received by the object using the force equation above.
b) What is the final momentum of the object at the end of the time interval?
c) What is the average net force exerted on the object over the time interval?
\vec{p} = m\vec{v}
\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}
\vec{J} = \int_{t_i}^{t_f}\vec{F}(t)\,dt
F_{\text{avg}} = \frac{J}{\Delta t}
a)So \int_{0}^{3.0}\vec{F}(t)\,dt = \left[ \left(t^3 -2.0t^2+10.0t \right)\hat{i} \right]_{0}^{3.0} = 39 \text{kg}\cdot\text{m}/\text{s}
b)This is where I was confused.
Well, since \vec{p} = m\vec{v}, I solved the indefinate integral
\int\vec{F}(t)\,dt = t^3-2t^2+10t + C
And since
\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} and \vec{p} = m\vec{v}
then
\vec{p}(0) = t^3 - 2.0t^2 + 10.0t + C = m\vec{v}_0 = 800,
so C = 800?
So final momentum is
\vec{p}(3) = (3)^3 -2(3)^2 + 10(3) + 800 = 815 \text{kg}\cdot\text{m}/\text{s}??
c)
F_{\text{avg}} = \frac{39}{3.0-0} = 13\,\text{N}
Thanks.
Homework Statement
A 40kg object with an initial velocity \vec{v}_0 = (20 \text{m}/\text{s})\hat{i} is accelerated by a varying net force \vec{F}(t)=\left[ \left( 3.0 \text{N}/\text{s}^2\right)t^2 - \left( 4.0 \text{N}/\text{s}^2 \right)t + 10.0\text{N} \right]\hat{i} over a time interval of 3.0 seconds.
a) What is the net implulse (\vec{J}) received by the object using the force equation above.
b) What is the final momentum of the object at the end of the time interval?
c) What is the average net force exerted on the object over the time interval?
Homework Equations
\vec{p} = m\vec{v}
\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}
\vec{J} = \int_{t_i}^{t_f}\vec{F}(t)\,dt
F_{\text{avg}} = \frac{J}{\Delta t}
The Attempt at a Solution
I had trouble at part b.a)So \int_{0}^{3.0}\vec{F}(t)\,dt = \left[ \left(t^3 -2.0t^2+10.0t \right)\hat{i} \right]_{0}^{3.0} = 39 \text{kg}\cdot\text{m}/\text{s}
b)This is where I was confused.
Well, since \vec{p} = m\vec{v}, I solved the indefinate integral
\int\vec{F}(t)\,dt = t^3-2t^2+10t + C
And since
\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} and \vec{p} = m\vec{v}
then
\vec{p}(0) = t^3 - 2.0t^2 + 10.0t + C = m\vec{v}_0 = 800,
so C = 800?
So final momentum is
\vec{p}(3) = (3)^3 -2(3)^2 + 10(3) + 800 = 815 \text{kg}\cdot\text{m}/\text{s}??
c)
F_{\text{avg}} = \frac{39}{3.0-0} = 13\,\text{N}
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