Physics, asked by bharatguptabst11, 8 months ago

A car of mass 500kg moving at a speed of 18 km/hr is stopped by applying breaks in 10 seconds. Calculate the force applied by the breaks.​

Answers

Answered by VishnuPriya2801
55

Answer:-

Given:

Mass of the car (m) = 500 kg

Initial speed / Velocity (u) = 18 km/h = 18 * 5/18 = 5 m/s

Final Velocity (v) = 0 m/s since the breaks are applied.

Time (t) = 10 s

We know that,

Force = mass * acceleration

And,

Acceleration = (v - u) / t

Hence,

Force (F) = 500 * (0 - 5) / 10

→ F = (500 * (- 5))/10

→ F = - 250 N

Hence, the force applied by the breaks is - 250 N.

Additional Information:

Force:

  • Force is the push or pull of an object.

  • It is a vector quantity.

  • It's S.I unit is Newton (N) and C.G.S is dyne (dyn) .

  • 1 N = 10⁵ dyn.
Answered by rocky200216
53

\large\mathcal{\orange{\overbrace{\green{\underbrace{\orange{SOLU}\green{TION}:-}}}}}

GIVEN :-

  • A car of mass 500kg moving at a speed of 18km/h is stopped by applying the brakes in 10 seconds .

✍️ So,

  • Mass (m) = 500kg

  • initial velocity (u) = 18 km/h

  • time (t) = 10 second

To Find :-

  • The force applied by the brakes .

CALCULATION :-

✍️ we have know that,

\bigstar\:\rm{\purple{\boxed{Force\:=\:Mass\:\times\:Acceleration\:}}}

________________________________

✍️ To calculate acceleration, we use the below equation

\bigstar\:\rm{\blue{\boxed{v\:=\:u\:+\:at}}}

✴️ Here,

  • v = final velocity = 0

  • u = initial velocity = 18 km/h = 5 m/s

  • t = time = 10 second

[NOTE :- final velocity is zero, because brakes are applied on it . So, at last car is stopped after sometime]

\rm{\implies\:0\:=\:5\:+\:a\:\times{10}\:}

\rm{\implies\:10\:a\:=\:-5\:}

\rm{\implies\:a\:=\:\dfrac{-5}{10}\:}

\rm{\green{\implies\:a\:=\:\dfrac{-1}{2}\:m/s^2}}

✍️ Now, the force applied on it is

\rm{\implies\:Force\:=\:500\times{\dfrac{-1}{2}}\:}

\bigstar\:\rm{\boxed{\pink{\implies\:Force\:=\:-250\:N\:}}}

[NOTE :- Here force is negative, because force is applied on the opposite direction of Motion]

✍️ Therefore, the force applied by the brakes is ‘250N’ .

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