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Sagot :
Answer:
[tex]\boxed {\boxed {\sf 115.15 \ J}}[/tex]
Explanation:
Kinetic energy is the energy an object possesses due to motion. It is calculated with the following formula.
[tex]E_K= \frac{1}{2} mv^2[/tex]
The mass of the object is 4.7 kilograms. The velocity of the object is 7 meters per second.
- m= 4.7 kg
- v= 7 m/s
Substitute the values into the formula.
[tex]E_K= \frac{1}{2} (4.7 \ kg)(7 \ m/s)^2[/tex]
Solve the exponent.
- (7 m/s)²= 7 m/s * 7 m/s = 49 m²/s²
[tex]E_K= \frac{1}{2} (4.7 \ kg)(49 \ m^2/s^2)[/tex]
Multiply the numbers together.
[tex]E_K = 2.35 \ kg * 49 \ m^2/s^2[/tex]
[tex]E_K= 115.15 \ kg*m^2/s^2[/tex]
Convert the units. 1 kilogram square meter per square second is equal to 1 Joule.
[tex]E_K= 115.15 \ J[/tex]
The object has 115.15 Joules of kinetic energy.
Explanation:
Kinetic energy, [tex]\textbf{\textit{K}}[/tex] is the energy associated with the state of motion of an object. This implies that as a particle accelerates by an external force, the kinetic energy possessed by the particle itself increases. Similarly, the kinetic energy of a particle decreases as an external force decelerates the particle and when the particle is stationary, it's kinetic energy is zero.
We account for such a change of kinetic energy as the energy transferred to the particle from an external force or from the particle to an external force, respectively. This transfer of energy, by a force [tex]\vec{F}[/tex], is defined as the work done, W on the particle by the force [tex]\vec{F}[/tex]. Positive work denotes the energy transferred to the particle whereas negative work is the transfer of energy from the particle.
Hence, it is obvious that the concept of work and kinetic energy are related and this relationship is defined by the work-energy theorem. The theorem states that that the work done, W by all forces acting on a particle (the resultant force) equals the change in the kinetic energy of the particle, formally written as
[tex]W \ = \ \Delta K \\ \\ W\ = \ K_{\text{final}} \ - \ K_{\text{initial}}[/tex]
Recall the kinematic equation for uniform acceleration,
[tex]{v_{\text{final}}}^{2} \ = \ {v_{\text{initial}}}^{2} \ + \ 2a\Delta x[/tex].
For a moving particle, [tex]\Delta x \ = \ d[/tex], where [tex]d[/tex] is the displacement of the particle, and from Newton's second law of motion, we know that the acceleration [tex]a \ = \ \displaystyle\frac{F}{m}[/tex]. Hence,
[tex]{v_{\text{final}}}^{2} \ = \ {v_{\text{initial}}}^{2} \ + \ \displaystyle\frac{2Fd}{m} \\ \\ \displaystyle\frac{1}{2}m{v_{\text{final}}}^{2} \ - \ \displaystyle\frac{1}{2}m{v_{\text{initial}}}^{2}\right) \ = \ Fd[/tex]
Additionally, work done, W on a particle by a constant force is the product of the component of the force in the direction of its displacement and the magnitude of the displacement.
[tex]W \ = \ Fd[/tex].
Thus,
[tex]W \ = \ \displaystyle\frac{1}{2}m{v_{\text{final}}}^{2} \ - \ \displaystyle\frac{1}{2}m{v_{\text{initial}}}^{2}\right)[/tex],
in which
[tex]K_{\text{final}} \ = \ \displaystyle\frac{1}{2}m{v_{\text{final}}}^{2}[/tex] and [tex]K_{\text{initial}} \ = \ \displaystyle\frac{1}{2}m{v_{\text{initial}}}^{2}[/tex].
Therefore, in general, for a particle of mass m moving with a speed v, the kinetic energy of the particle is defined as
[tex]K \ = \ \displaystyle\frac{1}{2}mv^{2}[/tex].
Substitute the known quantities into the expression above yields
[tex]K \ = \ \displaystyle\frac{1}{2}(4.7 \ \text{g})(7 \ \text{m s}^{-1})^{2} \\ \\ K \ = \ 115.15 \ \text{J}[/tex]
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