Question

An 1120 kg car traveling at 17.2 m/s is brought to a stop while skidding 40m. Calculate the work done on the car by the friction forces.

Answer 1

Work = 165670.4 J = 165.67 KJ

Step-by-step explanation:

First, we will find the deceleration of the car, using the 3rd equation of motion:

`2as = v_(f)^2 - v_(i)^2\\`

where,

a = deceleration = ?

s = skid distance = 40 m

vf = final speed = 0 m/s

vi = initial speed = 17.2 m/s

Therefore,

`2a(40\ m) = (0\ m/s)^2 - (17.2\ m/s)^2\\a = - 3.698\ m/s^2`

the negative sign indicates deceleration here.

Now, we will calculate the braking force applied by the brakes on the car:

`F = ma\\F = (1120\ kg)(-3.698\ m/s^2)\\F = - 4141.76\ N`

the negative sign indicates braking force.

Now, we will calculate the work done using the magnitude of this force:

`Work = |F|s\\Work = (4141.76\ N)(40\ m)\\`

Work = 165670.4 J = 165.67 KJ