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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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Most popular questions from this chapter

An accident victim with a broken leg is being placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and boot together have a mass of $4.0 k g$ , and the doctor has decided to hang a $6.0 k g$ mass from the rope. The boot is held suspended by the ropes, as shown in $F I G U R E P 6.41$ , and does not touch the bed.

a. Determine the amount of tension in the rope by using Newton鈥檚 laws to analyze the hanging mass.

b. The net traction force needs to pull straight out on the leg. What is the proper angle u for the upper rope?

c. What is the net traction force pulling on the leg?

Write a realistic problem for which these are the correct equations.
Draw the free-body diagram and the pictorial representation for your problem.
Finish the solution of the problem.

(100 N) \cos 鈦� 30^{鈭�} 鈭� f_{k} = (20 kg) a_{x}

$\begin{aligned} n + (100 N) \sin 鈦� 30^{鈭�} 鈭� (20 kg) (9.80 m / s^{2}) = 0 \\ f_{k} = 0.20 n . \end{aligned}$

An elevator, hanging from a single cable, moves upward at constant speed. Friction and air resistance are negligible. Is the tension in the cable greater than, less than, or equal to the gravitational force on the elevator? Explain. Include a free-body diagram as part of your explanation.

A ball tossed straight up has v = 0 at its highest point. Is it in equilibrium? Explain.

Problems 68 and 69 show a free-body diagram. For each:

a. Write a realistic dynamics problem for which this is the correct free-body diagram. Your problem should ask a question that can be answered with a value of position or velocity (such as "How far?" or "How fast?"), and should give sufficient information to allow a solution.

b. Solve your problem!

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