The work done is [tex]-1.36\cdot 10^5 J[/tex]
Explanation:
According to the work-energy theorem, the work done by external forces on the car is equal to the change in kinetic energy of the car. Therefore, we have:
[tex]W=\Delta K\\W = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
where
W is the work done by the external forces
m is the mass of the car
v is its final velocity
u is its initial velocity
In this problem, we have:
m = 1165 kg
[tex]u=55 km/h =15.3 m/s[/tex]
v = 0 (the car comes to a stop)
Solving for W, we find the work done by the frictional forces:
[tex]W=\frac{m(v^2-u^2)}{2}=\frac{(1165)(0-15.3^2)}{2}=-1.36\cdot 10^5 J[/tex]
Where the negative sign indicates that the direction of the force is opposite to the direction of motion of the car.
Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
#LearnwithBrainly
