Answer:
Part a)
all cars will travel equal distance before it stops
Part b)
Car F will have maximum work done by friction
Step-by-step explanation:
Part a)
As we know that the friction on Each car is given as
![F_f = \mu mg](https://img.qammunity.org/2020/formulas/physics/middle-school/8cerdfcxk4w0bnfjgs1kx9e0m0vye8swy2.png)
now the deceleration is given as
![a = -(F_f)/(m)](https://img.qammunity.org/2020/formulas/physics/high-school/uvlqx8mm3htj5gvr3y4x2ob8x3lwo1nftk.png)
![a = -\mu g](https://img.qammunity.org/2020/formulas/physics/high-school/52lzoqoq964x2coydn5qsuvjd729o1cczk.png)
so the deceleration is independent of the mass of the car
now the distance to stop the car is given as
![v_f^2 - v_i^2 = 2a d](https://img.qammunity.org/2020/formulas/physics/middle-school/1unr04sklyopzstgj6fzlo7g9r9nir8l80.png)
![0 - v^2 = -2(\mu g) d](https://img.qammunity.org/2020/formulas/physics/high-school/13e7at8eo86quenc7udj625cl6o03eatx1.png)
![d = (v^2)/(2\mu g)](https://img.qammunity.org/2020/formulas/physics/high-school/6igkf9dyf004lia9jjwf6l4ulkz0rtpsgu.png)
so all cars will travel equal distance before it stops
Part b)
Work done by friction force is given as
![W = F_f * d](https://img.qammunity.org/2020/formulas/physics/high-school/fyfu06ubny344ipnfj0q19yhnaga2tjqvt.png)
so we have
![W_f = \mu mg d](https://img.qammunity.org/2020/formulas/physics/high-school/8s9gm9mh0d2gfvmc6ot11hmsynpmp7fqqs.png)
so most massive car will have maximum work done done by friction
so Car F will have maximum work done by friction