Answer:
x = 0.105 meters
Explanation:
It is given that,
A box slides from rest down a friction less ramp inclined at 34.0° with respect to the horizontal, [tex]\theta=34^{\circ}[/tex]
Spring constant of the spring, [tex]k=3.5\times 10^4\ N/m[/tex]
Mass of the box, m = 12 kg
Th box slides 3 m from the point of release to the point where it comes to rest against the spring. We need to find the compression of the spring when the box comes to rest.
Let h is the height of the incline,
[tex]h=sin(34)\times 3=1.67\ m[/tex]
Using the conservation of energy in case of spring,
[tex]\dfrac{1}{2}kx^2=mgh[/tex]
[tex]x=\sqrt{\dfrac{2mgh}{k}}[/tex]
[tex]x=\sqrt{\dfrac{2\times 12\times 9.8\times 1.67}{3.5\times 10^4}}[/tex]
x = 0.105 meters
So, the compression of the spring is 0.105 meters. Hence, this is the required solution.
