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A .183 kg ball is moving 18.8 m/s when it runs into a spring of spring constant 86.9 N/m. How much KE does the ball have when it has compressed the spring 0.520 m? (Unit=J)

User Cloned
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1 Answer

5 votes

Answer:

The value is
KE_B = 20.59 \ J

Step-by-step explanation:

From the question we are told that

The mass of the ball is
m = 183 \ kg

The initial speed of the ball is
u = 18.8 \ m/s

The spring constant is
k = 86.9 \ N/m

The compression distance is
x = 0.520 \ m

Generally the energy stored in the string is mathematically represented as


E = (1)/(2) * k * x^2

=>
E = (1)/(2) * 86.9 * 0.520 ^2

=>
E = 11.75 \ J

Generally the kinetic energy of the ball is mathematically represented as


KE_b = (1)/(2) * m * u^2

=>
KE_b = (1)/(2) * 0.183 * (18.8 )^2


KE_b = 32.34 \ J

Generally the KE the ball have when it has compressed the spring is mathematically represented as


KE_B = KE_b - E

=>
KE_B = 32.34 - 11.75

=>
KE_B = 20.59 \ J

User Mgsloan
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