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madreJ [45]
3 years ago
9

An athlete stretches a spring an extra 40.0 cm beyond its initial length. How much energy has he transferred to the spring, if t

he spring constant is 52.9 N/cm?
a) 4230 kJ
b) 4230 J
c) 423 kJ
d) 423 J
Physics
1 answer:
notsponge [240]3 years ago
4 0

Answer:

<h2>b) 4230 J </h2>

Explanation:

Step one:

given data

extension= 40cm

Spring constant K= 52.9N/cm

Step two:

Required

the Kinetic Energy KE

the expression to find the kinetic energy is

KE= 1/2ke^2

substituting our data we have

KE= 1/2*52.9*40^2

KE=0.5*52.9*1600

KE= 42320Joules

<u>The answer is b) 4230 J </u>

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Marina CMI [18]
Picture #1:
GPE = (mass) x (gravity) x (height)
GPE = (2 kg) x (9.8 m/s²) x (40 m) = 784 joules

KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (5 m/s)²
KE = (1 kg) (25 m²/s²)  =  25 joules

Picture #2:
KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (10 m/s)²
KE = (1 kg) (100 m²/s²)  =  100 joules

Picture #3:
GPE = (mass) x (gravity) x (height)
GPE = (20 kg) x (9.8 m/s²) x (2 m) = 392 joules

KE = (1/2) (mass) (speed²)
KE = (1/2) (20 kg) (5 m/s)²
KE = (10 kg) (25 m²/s²)  =  250 joules

Picture #4:
GPE = (mass) x (gravity) x (height)
98 joules = (1 kg) x (9.8 m/s²) x (height)
Height = (98 joules) / (1 kg x 9.8 m/s²)
Height = 10 meters

Picture #5:
GPE = (mass) x (gravity) x (height)
39,200 Joules = (mass) x (9.8 m/s²) x (20 m)
Mass = (39,200 joules) / (9.8 m/s² x 20 m)
Mass = 200 kg

5 0
3 years ago
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