The magnitude of the work done by friction in stopping the car 312500 J
v² = u² + 2as
0² = 25² + (2 × a × 10)
0 = 625 + 20a
Collect like terms
0 – 625 = 20a
–625 = 20a
Divide both side by 20
a = –625 / 20
a = –31.25 m/s²
NOTE: The negative sign indicates that the car is coming to rest (i.e decelerating)
F = ma
F = 1000 × 31.25
F = 31250 N
Wd = F × d
Wd = 31250 × 10
Wd = 312500 J
Learn more about workdone:
https://brainly.com/question/26239935