Question

What would be the force required to stop car of mass 1000kg and a loaded truck of mass 10,00kg in 2 second each moving with velocity 5m/s

Answer 1

Answer:-

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Given:-

$\implies \sf Initial\: Velocity = 5\dfrac{m}{s} \\ \implies \sf Final \: Velocity = 0\dfrac{m}{s} \\ \implies \sf Time \: interval = 2second$

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To find:-

$\implies \sf Acceleration (?)$

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Solution:-

We know that:-

$\longrightarrow \sf Force = mass \times acceleration.$

or

$\longrightarrow \sf Force = mass \times deacceleration$

And

$\sf Acceleration =\dfrac{Final \: velocity - Initial \: Velocity}{Time \: Interval}$

So, here from given above the acceleration will be:-

$\sf \therefore Acceleration = \dfrac{0 - 5}{2} = 2.5m/s^{2}$

And deacceleration will be:-

$\sf -2.5m/s^{2}$

[As acceleration = -deacceleration]

Now for car,

Mass = 1000kg

Force = mass $\times$ deacceleration

= 1000$\times$ 2.5m/s²

= 2500N.

Now for truck,

Mass = 10,000kg

Force= 10,000 $\times$ (2.5)

= 25000N.

Hence, Solved.

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Some formulas:-

$\mapsto \sf s = ut + \dfrac{1}{2} at^{2} \\ \\ \mapsto \sf v = u + at \\ \\ \mapsto \sf 2as = v^{2} - u^{2} \\ \\ \mapsto \sf p = m \times v \\ \\ \mapsto \sf a = \dfrac{v - u}{t} \\ \\ \mapsto \sf Speed = \dfrac{distance}{time}$