Astronauts in space "weigh" themselves by oscillating on a CALC spring. Suppose the position of an oscillating 75kgastronaut is given byx=(0.30m)sin((πrad/s)×t), where tis in s. What force does the spring exert on the astronaut at

(a) t=1.0sand

(b) 1.5s ? Note that the angle of the sine function is in radians.

Short Answer

Expert verified

Part(a) Force at time t=1sisF=0N.

Part (b) Force at,=1.5sisF=225N2.

Step by step solution

01

Step(a) Step 1: Given.

Mass of astronaut =75kg.

Displacement,x=0.31sinπt

02

Part(a) Step 2: Formula used.

The formula of force is given as:

F=ma

03

Part (a) Step 3: Calculation.

The equation of displacement is given as:

x=0.31sin(π)tv=dxdt=(0.31)πcosπta=dvd=0.31×(π)2sin(π)t

Force At t=1s

a=0.31π2sinπ×I=0m/s2

Since,

F=maF=75×0=0N

04

Part (b) Step 4: Given.

Mass of astronaut =75kg.

Displacement,x=0.31sinπt.

05

Part (b) Step 5; Formula Used.

The formula of force is given as:

F=ma

06

Part (b) Step 6: Calculation.

The equation of displacement is given as:

x=0.31sin(π)tv=dxdt=(0.31)πcosπta=dyd=0.31×(π)2sin(π)t

Force At t=1.5s

a=0.31π2sinπ×1.5=3m/s2

Since,

F=maF=75×3=225N

07

Part (b) Step 7: Conclusion.

Part (a): Force at timet=1sisF=0N.

Part (b): Force At, t=1.5sis 225N.

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