Question

What is the impulse on a $3.0kg$particle that experiences the force shown in FIGURE EX11.4?

Short answer

The impulse is$J_x=6N\times s$

Step-by-step solution

Given information  

We need to find the impulse on the particle that experiences th force.

Simplify  

A force for a short time when a object recives is called impulse. it is the area under the curve of the force-versus-time graph and it is the same as the change in momentum. The impulse is the quantity $j_x$and it is given by equation in the form

$$\begin{gathered} impulse=J_x={\int }_{t_i}^{t_f}{F}_x\left(t\right)dt \\ =areaunderthe{F}_x\left(t\right)curvebetween{t}_{i}and{t}_f\left(1\right) \end{gathered}$$

The graph we have is force versus time in figure, so the force is between the range of time from $t_i=2msto{t}_f=8ms$. the impulse take the shape of triangle and the height is $h=2000N$and the base is $(b=8ms-2ms=6ms)$. the area of triangle is half product of the height and the base.

$A=\frac{1}{2}\left(height\right)\left(base\right)$

To get impulse we can use equation $\left(1\right)$

localid="1649418166522" $$\begin{gathered} J_x=\frac{1}{2}\left(height\right)\left(base\right) \\ =\frac{1}{2}\left(2000N\right)(6\times {10}^{-3}s) \\ =6N\times s \end{gathered}$$