Answer
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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.
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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