Question

A block of mass 11.0 kg slides from rest down a frictionless 38.0° incline and is stopped by a strong spring with

Answer 1

Answer:

11.72 mm

Explanation:

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

$PE_{gravitational}=PE_{spring}$

$mgh=0.5kx^{2}$ 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

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

Making x the subject then

$x=\sqrt {\frac {2mgdsin\theta}{k}}$

Substituting the given values then

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