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The question asks to find the work done by the force $\mathbf{F(t)}=(4\mathbf{i}+12t^2\mathbf{j})N$ in the time interval $0\leq{t}\leq{1}$ on a particle of mass $4kg$, however I am not sure my working is correct, to start with I found $\mathbf{a(t)}$ by writing $\mathbf{F(t)}$ as $4\mathbf{a(t)}$, therefore:

$$\mathbf{a(t)}=(\mathbf{i}+3t^2\mathbf{j})ms^{-2}$$ $$\therefore \mathbf{v(t)}=\int_{0}^{t}(\mathbf{i}+3t^2\mathbf{j})ms^{-2}dt$$ $$\therefore \mathbf{v(t)}=(t\mathbf{i}+t^3\mathbf{j})ms^{-1}$$

Therefore the work done by the force is given by: $$W=\int_{0}^{1}\mathbf{F(t)}\cdot dx$$ $$=\int_{0}^{1}\mathbf{F(t)}\cdot \mathbf{v(t)}dt$$ $$=\int_{0}^{1}(4\mathbf{i}+12t^2\mathbf{j})N\cdot(t\mathbf{i}+t^3\mathbf{j})ms^{-1}dt$$ $$=\int_{0}^{1}4t+12t^5dt$$ $$=\left[2t^2+2t^6\right ]_0^1$$ $$=4J$$

Is this correct? Thanks

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closed as off-topic by David Z Oct 30 '13 at 19:48

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    $\begingroup$ Looks good to me. Although you should place a space (latex: \,) between the number and unit. $\endgroup$ – Ignacio Vergara Kausel Oct 30 '13 at 16:32
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    $\begingroup$ And you should use a space and roman characters for the differential like $\ldots\,{\rm d}t$ and for units $\ldots\,{\rm m s^{-1}}$. $\endgroup$ – ja72 Oct 30 '13 at 17:04
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It is correct on the condition that $\mathbf{v}(0)=\mathbf{0}$. $$\int_0^t \mathbf{a}(t) dt =\mathbf{v}(t)-\mathbf{v}(0).$$

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