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Alinara [238K]
3 years ago
5

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force)

acts on her. The student has a mass of 61.0 kg, and the height of the water slide is 12.3 m. If the kinetic frictional force does -5.80 × 103 J of work, how fast is the student going at the bottom of the slide?
Physics
1 answer:
katen-ka-za [31]3 years ago
7 0

Answer:

Velocity will be equal to 7.31 m/sec

Explanation:

We have given mass of the student m = 61 kg

Height of the water slide h = 12.3 m

Acceleration due to gravity g=9.8m/sec^2

Potential energy is equal to U=mgh=61\times 9.8\times 12.3=7352.94J

Work done due to friction = -5800 J

So energy remained = 7352.94-5800 = 1552.94 J

This energy will be equal to kinetic energy

So \frac{1}{2}mv^2=1552.94

\frac{1}{2}\times 61\times v^2=1552.94

v^2=50.91

v = 7.13 m/sec

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A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
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Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

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Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

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