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3 years ago

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A single force acts on a 2.0 kg particle-like object in such a way that the position of the object as a function of time is give

n by x = 3.0t − 4.0t 2 + 1.0t 3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 5.0 s.

1 answer:

kaheart [24]3 years ago

8 0

Answer:

The work done by the force is 1435 Joule.

Explanation:

You have to apply the work-energy theorem, which associates the work and the mechanical energy

W of all non conservative forces = ΔEm = Emb - Ema

where Emb is the mechanical energy in point b and Ema is the mechanical energy in point a

Is also known that mechanical energy is the sum of potential energy and kinetic energy.

In this case, there is only kinetic energy because the potential energy is given by conservative forces like the weight but the weight is perpendicular to the displacement and because of that it doesn't produce work.

So, the work is given by:

W= Ekb - Eka, where Ek is the kinetic energy

By definition the kinetic energy is given by:

$Ek=\frac{1}{2}mV^{2}$ where m is the mass and V is the speed.

Replacing the definition of Ek in the equation of work:

$W=\frac{1}{2}mVb^{2} - \frac{1}{2}mVa^{2}$

So, you have to calculate the speed in point a and in point b

You can calculate the velocity of the object as a function of time derivating the equation of the position:

$V(t) = \frac{dX(t)}{dt} = \frac{d(3.0t-4.0t^{2}+1.0t^{3}) }{dt}$

$V(t)=3.0-8.0t+3.0t^{2}$

Replacing t=0 s to obtain the speed in the initial point (point a)

Va=3.0-8(0)+3.0(0)(0)

Va=3.0 m/s

Replacing t=5.0 to obtaing the speed in final point (point b)

Vb= 3-8(5)+3(5)(5) = 38 m/s

Therefore the work is:

$W = \frac{1}{2}(2.0)(38)^{2} - \frac{1}{2}(2.0)(3^{2}) =1444 - 9 = 1435 [J]$

<u>Where J is Joules, the unit for work.</u>

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Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

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Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

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$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

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