Answer:
Explanation:
Let the frictional upward force be F = 550 N
Net force on it = mg - F
82 X 9.8 - 550
= 803.6 - 550
253.6 N
Downward acceleration = 253.6 / 82
= 3.0927 m / s²
final velocity after slipping 5.2 m , v then
v² = 2 x a s
= 2 x 3.0927 x 5.2
v = 5.67 m /s
It stops in .45 m distance
v² = u² - 2as
0 = 5.67² - 2 a x .45
a = - 35.72 m / s²