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Chapter 6: Q. 61 (page 157)

Astronauts in space "weigh" themselves by oscillating on a CALC spring. Suppose the position of an oscillating $75 kg$ astronaut is given by $x = (0.30 m) \sin 鈦� ((蟺 rad / s) 脳 t)$ , where $t$ is in $s$ . What force does the spring exert on the astronaut at
(a) $t = 1.0 s$ and
(b) $1.5 s$ ? Note that the angle of the sine function is in radians.

Short Answer

Expert verified

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

Part (b) Force at, $= 1 . 5_{s}$ is $F = 225 N_{2}$ .

Step by step solution

01

Step(a) Step 1: Given.

Mass of astronaut $= 75 kg$ .

Displacement, $x = 0.31 \sin 鈦� 蟺 t$

02

Part(a) Step 2: Formula used.

The formula of force is given as:

$F = m a$

03

Part (a) Step 3: Calculation.

The equation of displacement is given as:

$\begin{aligned} x = 0.31 \sin 鈦� (蟺) t \\ v = \frac{d x}{d t} = (0.31) 蟺 \cos 鈦� 蟺 t \\ a = \frac{d v}{d} = 鈭� 0.31 脳 (蟺)^{2} \sin 鈦� (蟺) t \end{aligned}$

Force At $t = 1 s$

$\begin{matrix} a = 鈭� 0.31 蟺^{2} \sin 鈦� 蟺脳 I \\ = 0 m / s^{2} \end{matrix}$

Since,

$\begin{aligned} F = m a \\ F = 75 脳 0 \\ = 0 N \end{aligned}$

04

Part (b) Step 4: Given.

Mass of astronaut $= 75 kg$ .

Displacement, $x = 0.31 \sin 鈦� 蟺 t$ .

05

Part (b) Step 5; Formula Used.

The formula of force is given as:

$F = m a$

06

Part (b) Step 6: Calculation.

The equation of displacement is given as:

$\begin{aligned} x = 0.31 \sin 鈦� (蟺) t \\ v = \frac{d x}{d t} = (0.31) 蟺 \cos 鈦� 蟺 t \\ a = \frac{d y}{d} = 鈭� 0.31 脳 (蟺)^{2} \sin 鈦� (蟺) t \end{aligned}$

Force At $t = 1.5 s$

$\begin{matrix} a = 鈭� {0.31}_{蟺}^{2} \sin 鈦� 蟺脳 1.5 \\ = 鈭� 3 m / s^{2} \end{matrix}$

Since,

$\begin{aligned} F = m a \\ \begin{aligned} F & = 75 脳 3 \\ = 225 N \end{aligned} \end{aligned}$

07

Part (b) Step 7: Conclusion.

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

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

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