Question
Asked Nov 12, 2019
When a flea (m = 600 μg) is jumping up, it extends its legs 0.5 mm and reaches a speed of 0.50 m/s in that time. How much work did the flea do during that time? Use g = 10 m/s2
 
 
 
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Expert Answer

Step 1

According to the kinematics equation, find the acceleration of the flea.

Substitute the values,

2-u22as
a =
2s
(0.50m/s)- (0m/s)
0.001m
2(0.5mm)
1mm
250m/s2
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2-u22as a = 2s (0.50m/s)- (0m/s) 0.001m 2(0.5mm) 1mm 250m/s2

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Step 2

The net force acting on the flea is,

Substitute the values,

F-mg ma
F, 3D та+ mg
m(a+g)
flea
10 kg
(250m/s210m/s2
ug
Fon (600ug)
ea
(600x10kg) (260m/s)
31.56 х10*N
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Image Transcriptionclose

F-mg ma F, 3D та+ mg m(a+g) flea 10 kg (250m/s210m/s2 ug Fon (600ug) ea (600x10kg) (260m/s) 31.56 х10*N

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Step 3

The work done by the flea ...

W = Fd
103m
(1.56x10 N)(0.5mm)
mm
106 J
J
1J
0.078 x 10
0.078 J
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W = Fd 103m (1.56x10 N)(0.5mm) mm 106 J J 1J 0.078 x 10 0.078 J

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Physics

Kinematics

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