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**Hint:**In order to answer this question we should first get some idea about force. A force is an external agent that has the ability to change the state of rest or motion of a body. There is a magnitude to it, as well as a direction to it. The force's location is known as the force's direction, and the force's application point is known as the force's application point.

**Complete step by step answer:**

Now let us understand about the Centripetal Force. A centripetal force is a force that causes a body to follow a curved path (from Latin centrum, "centre," and peter, "to seek"). It always travels in the opposite direction of the body, toward the instantaneous centre of curvature of the road.

A force by which bodies are drawn, impelled, or in some other way tend, towards a point such as a centre," wrote Isaac Newton. Gravity produces the centripetal force that causes celestial orbits in Newtonian mechanics. Any object moving with velocity \[v\] along a circular path of radius\[\;r\] experiences an acceleration directed toward the path's core is known as centripetal acceleration.

$a = \dfrac{{{v^2}}}{r}$

As we know that,

$F = ma$

$\Rightarrow a = \dfrac{F}{m}$

Equating this to centripetal force:

$\dfrac{{{v^2}}}{r} = \dfrac{F}{m}$

Now we can say that by see the above equation that the centripetal force ${F_c}$ has magnitude

${F_c} = \dfrac{{{v^2}m}}{r}$

Here, $v = velocity$, $r = radius$ and $m = mass$.

Given :

$v = 36\,km/h = 36 \times \dfrac{5}{{18}} = 10\,m/s \\

\Rightarrow r = 50m \\

\Rightarrow m = 1000\,kg \\$

$\Rightarrow {F_c} = \dfrac{{m{v^2}}}{r} \\

\Rightarrow {F_c} = \dfrac{{1000 \times {{10}^2}}}{{50}} \\

\therefore {F_c}= 2 \times {10^3}\,N$

**Hence, the required centripetal force is $2 \times {10^3}\,N$.**

**Note:**The force that induces circular motion is known as the centripetal force. Even if an object moves at a constant pace, a force must still act on it as it moves in a circle. The object would' shoot off' in a direction tangential to the circle if it didn't exist.

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