Question

Astronauts in space "weigh" themselves by oscillating on a CALC spring. Suppose the position of an oscillating $75\mathrm{kg}$astronaut is given by$x=\left(0.30\mathrm{m}\right)\sin \left(\right(\pi \mathrm{rad}/\mathrm{s})\times t)$, where $t$is in $\mathrm{s}$. What force does the spring exert on the astronaut at

(a) $t=1.0\mathrm{s}$and

(b) $1.5\mathrm{s}$ ? Note that the angle of the sine function is in radians.

Short answer

Part(a) Force at time $t=1_{s}isF=0N.$

Part (b) Force at,$=1.5_s$is$F=225N_2$.

Step-by-step solution

Step(a) Step 1: Given.

Mass of astronaut $=75\mathrm{kg}$.

Displacement,$x=0.31\sin \pi t$

Part(a) Step 2: Formula used.

The formula of force is given as:

$F=ma$

Part (a) Step 3: Calculation.

The equation of displacement is given as:

$\begin{array}{r}\mathrm{x}=0.31\sin \left(\pi \right)t\\ \mathrm{v}=\frac{dx}{dt}=\left(0.31\right)\pi \cos \pi \mathrm{t}\\ a=\frac{dv}{d}=-0.31\times (\pi {)}^2\sin (\pi )t\end{array}$

Force At $\mathrm{t}=1\mathrm{s}$

$\begin{array}{c}a=-0.31{\pi }^2\sin \pi \times I\\ =0\mathrm{m}/{\mathrm{s}}^2\end{array}$

Since,

$\begin{array}{r}F=ma\\ F=75\times 0\\ =0\mathrm{N}\end{array}$

Part (b) Step 4: Given.

Mass of astronaut $=75\mathrm{kg}$.

Displacement,$x=0.31\sin \pi t$.

Part (b) Step 5; Formula Used.

The formula of force is given as:

$F=ma$

Part (b) Step 6: Calculation.

The equation of displacement is given as:

$\begin{array}{r}\mathrm{x}=0.31\sin \left(\pi \right)t\\ \mathrm{v}=\frac{dx}{dt}=\left(0.31\right)\pi \cos \pi \mathrm{t}\\ a=\frac{dy}{d}=-0.31\times (\pi {)}^2\sin (\pi )\mathrm{t}\end{array}$

Force At $\mathrm{t}=1.5\mathrm{s}$

$\begin{array}{c}a=-{0.31}_{\pi }^2\sin \pi \times 1.5\\ =-3\mathrm{m}/{\mathrm{s}}^2\end{array}$

Since,

$\begin{array}{r}F=ma\\ \begin{array}{rl}F& =75\times 3\\ & =225\mathrm{N}\end{array}\end{array}$

Part (b) Step 7: Conclusion.

Part (a): Force at time$t=1_s$is$F=0N.$

Part (b): Force At, $t=1.5s$is $225N.$