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zhannawk [14.2K]
3 years ago
8

If a bow holds 500J of potential energy as the arrow is pulled back, how much kinetic energy will the arrow have after it has be

en released?
Physics
1 answer:
ale4655 [162]3 years ago
4 0

Answer:

500J

Explanation:

The arrow will have an energy of 500J after it has been released from its state of rest.

This is compliance with the law of conservation of energy which states that "in every system, energy is neither created nor destroyed but transformed from one form to another".

  • The energy at rest which is the potential energy is 500J
  • This energy will be converted to kinetic energy in total after the arrow has been released.
  • This way, no energy is lost and we can account for the energy transformations occurring.
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Answer:

the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

Explanation:

Given that,

mass, m = 1kg

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By Newtons second law ,

The diffrential equation of motion with damping is given by

m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}

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1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}

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suppose the equation of the form x =e^m^t,

and the auxilliary equation is given by

m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3

The general solution for the above differential equation is

x(t) =C_1e^{-3t}+C_2e^{-7t}

Derivate with respect to t

x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

x'(0) = 0

substitute the initial condition

C_1 +C_2 = 1

-3C_1-7C_2=0

Solve the above equation to get C₁ and C₂

C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}

substitute for C₁ and C₂ in general solution

x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

Thus the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

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