A block of mass 11.0 kg slides from rest down a frictionless 38.0° incline and is stopped by a strong spring with
k = 2.90 ✕ 104 N/m.
The block slides 3.00 m from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed?

Relax

Respuesta :

Answer:

11.72 mm

Explanation:

The gravitational potential energy equals the potential energy of the spring hence

[tex]PE_{gravitational}=PE_{spring}[/tex]

[tex]mgh=0.5kx^{2}[/tex] where m is the mass of object, g is the acceleration due to gravity, h is the height, k is the spring constant and x is the extension of the spring

[tex]mgdsin\theta=0.5kx^{2}[/tex] where \theta is the angle of inclination and d is the sliding distance

Making x the subject then

[tex]x=\sqrt {\frac {2mgdsin\theta}{k}}[/tex]

Substituting the given values then

[tex]x=\sqrt{\frac {2\times 11\times 9.81\times 3\times sin 38}{2.9\times 10^{4}}}= 0.117240716\approx 11.72 mm[/tex]