Question

Force $F\left(x\right)=\left(10N\right)\sin \left(\raisebox{1ex}{$2\pi t$}\!\left/ \!\raisebox{-1ex}{$4s$}\right.\right)$is exerted on a 250 g particle during the interval$0s\le t\le 2.0s$. If the particle starts from rest, what is its speed at t = 2.0 s?

Short answer

The speed at$t=2.0sec$is$50.92m/s$

Step-by-step solution

Step 1. Given information

Force expression, $F\left(x\right)=\left(10N\right)\sin \left(\raisebox{1ex}{$2\pi t$}\!\left/ \!\raisebox{-1ex}{$4s$}\right.\right)$

The time variation of force value.

Step 2. Apply and calculate

We apply the impulse-momentum theorem,

$$\begin{gathered} \begin{array}{l}{\int }_0^2\left(10N\right)\sin \left(\frac{2\pi t}{4}\right)=0.25\times v_{\text{final}}\\ -\frac{4}{2\pi }\left(\text{cos}\pi -\text{cos}0\right)\times 10=0.25\times v_{final}\end{array} \\ {v}_{final}=50.92m/s \end{gathered}$$