# Calculate the force needed to bring a 850 kg car to rest from a speed of 90.0 km/h in a distance...

## Question:

Calculate the force needed to bring a {eq}\rm 850\ kg {/eq} car to rest from a speed of {eq}\rm 90.0\ km/h {/eq} in a distance of {eq}\rm 105\ m {/eq} (a fairly typical distance for a nonpanic stop).

## Kinetic Energy:

When an object is moving from one place to another place, then it possesses some energy due to its movement, this energy is known as the kinetic energy of the object. As the velocity of the object increases, the value of the kinetic energy also increases.

Given data:

• Mass of the car, {eq}m = 850\;{\rm{kg}} {/eq}
• Final velocity of the car, {eq}{v_f} = 90.0\;{\rm{km/h}} = 25\;{\rm{m/s}} {/eq}
• Distance, {eq}d = 105\;{\rm{m}} {/eq}

From work energy theorem,

{eq}W = \Delta K = {K_f} - {K_i}.......................(1) {/eq}

Here,

Work done is, {eq}W = F \times d {/eq}

Final kinetic energy of the car, {eq}{K_f} = \dfrac{1}{2}mv_f^2 {/eq}

Initial kinetic energy of the car, {eq}{K_i} = \dfrac{1}{2}mv_i^2 {/eq}

Now, substitute these values in equation (1)

{eq}F \times d = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2 {/eq}

Since, the car starts from rest so its initial velocity will be, {eq}{v_i} = 0 {/eq}

Substitute the given values,

{eq}\begin{align*} F \times 105& = \dfrac{1}{2} \times 850 \times {\left( {25} \right)^2}\\ F &= \dfrac{{265625}}{{105}}\\ F &= 2529.76\;{\rm{N}} \simeq {\rm{2}}{\rm{.52}} \times {\rm{1}}{{\rm{0}}^3}\;{\rm{N}} \end{align*} {/eq}

Therefore, the required force is {eq}{\rm{2}}{\rm{.52}} \times {\rm{1}}{{\rm{0}}^3}\;{\rm{N}} {/eq} .