The 2-Mg car has a velocity of v1 = 100 km/h when the driver sees an obstacle in front of...

Free-body diagram of car:

Let \(F_{f}\) is the frictional force, which is acting opposite to the motion of the car

$$ F_{f}=\mu_{k} N $$

Equations of motion:

$$ \begin{aligned} &+\uparrow \sum F_{y}=m a_{y} \\ &N-W=m a_{y} \\ &N-m g=m(0) \\ &N=m g \\ &N=2000 \times 9.81 \\ &N=19620 \mathrm{~N} \end{aligned} $$

Substituting value of \(\mathrm{N}\) in \(F_{f}=\mu_{k} N\)

$$ \begin{aligned} &F_{f}=\mu_{h} \times 19620 \\ &F_{f}=19620 \mu_{k} \end{aligned} $$

A force does work when it moves through a displacement in the direction of the force. Here frictional force is acting in opposite direction to the displacement of work. So \(F_{f}\) does negative work. Normal force \(N\), weight w does no work since they never undergoes displacement along its line of action.

Principle of work and energy:

Applying the work-energy principle

$$ T_{1}+\sum U_{1-2}=T_{2} \quad--(1) $$

Where, \(T_{1}\) - initial kinetic energy

\(T_{2}-\) Final kinetic energy

\(\sum U_{1-2}-\) Work done by all forces

Substitute the known values in equation (1)

$$ \frac{1}{2} m v_{1}^{2}-\left(F_{f} \times d^{\prime}\right)=\frac{1}{2} m v_{2}^{2} $$

Where, \(d^{\prime}\) is the skidding distance of the car

$$ \begin{aligned} &\left(\frac{1}{2} \times 2000 \times 27.778^{2}\right)-\left(19620 \mu_{k} \times d^{\prime}\right)=\frac{1}{2} m \times 0^{2} \\ &19620 \mu_{k} d^{\prime}=771617.284 \\ &d^{\prime}=\frac{39.328}{\mu_{k}} \end{aligned} $$

Car's brakes are applied after \(0.75 \mathrm{~s}\)

The distance traveled by the car during the reaction time, \(d^{\prime \prime}=v_{1} t\)

$$ \begin{aligned} &d^{\prime \prime}=27.778 \times 0.75 \\ &d^{\prime \prime}=20.83 \mathrm{~m} \end{aligned} $$

Thus the total distance traveled by car before it stops is \(d=d^{\prime}+d^{*}\)

$$ \begin{aligned} &175=\frac{39.327}{\mu_{k}}+20.83 \\ &\frac{39.327}{\mu_{k}}=154.17 \\ &\mu_{k}=\frac{39.327}{154.17} \\ &\mu_{k}=0.255 \end{aligned} $$