Answer:
Step-by-step explanation:
Frictional force acting on the box = mg cos 25 x μ where μ is coefficient of friction
= 2.4 x 9.8 x cos 25 x .51 = 10.87 N
Weight of the box acting along the incline = mg sin 25 = 9.94 N
Net force = 10.87 - 9.94 = .93 N . It will create a deceleration of .93 / 2.4
= .3875 m /s² in the box.
v² = u² - 2 a s
v = 0 , a = .3875 , s = 5.4 m
= = u² - 2 x .3875 x 5.4
u = 2.045 m /s