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VARVARA [1.3K]
3 years ago
9

What is the deflection of a spring? A ball of mass 2kg is dropped from a height if 20cm on a spring of stiffness k=1225 N/m. Fin

d the maximum deflection of the spring.
Physics
1 answer:
ollegr [7]3 years ago
6 0
Use the law of conservation of energy. Potential energy of the object with mass m at the height h is equal to:

<span> E=mgh

</span> and the potential energy of a spring is equal to <span> <span>E<span>spring</span></span>=<span>1/2</span>k<span>x^2</span></span><span> where k = spring constant, x = displacement Since all the gravitational potential energy will be transformed to the spring potential energy, its quite easy to figure out the displacement from this two equations 

E= mgh
E= 2*9.8*0.2
E= 3.92

E=1/2kx^2
E=1/2*1225*X^2
3.92*2=1225*X^2
0.0064=X^2
X=0.08 m or 8 cm</span>
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Answer:

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How do conservation tillage practices lead to agricultural sustainability?
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James and Juan were at the highest point in the football stadium, dropping water balloons on people below. Many of the balloons
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Read 2 more answers
A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air res
Andre45 [30]

Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

mass of golf ball m = 46.8 g = 0.0468 kg

initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

so

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef  = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

v₂ = √( 3457.44 - 325.164 )  

v₂ = √( 3132.276 )

v₂ = 55.97 m/s

Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

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