It was about 9:30 p.m. sorry if the answer is wrong
Answer:
The maximum height the box will reach is 1.72 m
Explanation:
F = k·x
Where
F = Force of the spring
k = The spring constant = 300 N/m
x = Spring compression or stretch = 0.15 m
Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N
Mass of box = 0.2 kg
Work, W, done by the spring =
and the kinetic energy gained by the box is given by KE = 
Since work done by the spring = kinetic energy gained by the box we have
=
therefore we have v =
=
=
= 5.81 m/s
Therefore the maximum height is given by
v² = 2·g·h or h =
=
= 1.72 m
Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :

Horizontal component of force(Normal reaction) :

Since, box is on the verge of slipping :

Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.