A particle moving in the x direction is being acted upon by a net force f(x)=cx2, for some constant

1 answer:

6 0

The work-energy theorem states that the change in kinetic energy of the particle is equal to the work done on the particle:
$\Delta K = W$
The work done on the particle is the integral of the force on dx:
$W= \int\limits^{3L}_L {F(x)} \, dx = \int\limits^{3L}_L {cx^2} \, dx = \frac{26}{3}cL^3$
So, this corresponds to the change in kinetic energy of the particle.

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<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider $M_{1}$ and $M_{2}$ as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

$\text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}$

As gravitational constant $G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}$ , $M_{1}$ = 20 kg and $M_{2}$ = 100 kg, while d = 2.6 m, then