Question

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 39.0 ∘ above the horizontal. The glider has mass 5.00×10−2 kg. The spring has 600 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.50 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. What is the kinetic energy of the glider at this point?

Answer 1

Answer:

kinetic energy of glider = -0.376 j      

Explanation:

Original distance by which the spring is compressed   can be determine by using law of  conservation of energy (As spring slide from glider, it gain some  potential energy.  On the top of track there is glider’s potential energy exist)

$\frac{1}{2} kx^2 = mgh$

where x is compressed distance by spring

height of track is calculated as

$sin\theta \frac{h}{l}$

$h = l sin\theta$

where l is distance covered by glider, theta angle is the angle of inclination of track

therefore, we have formula for x as

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

$x = \sqrt{\frac{2\times 0.05 \times 9.8 \times 1.50\times sin39}{600}}$

x =  0.039 m

kinetic energy of glider  = potential energy by spring -  potential energy at 1.50 distance

                                     $=( \frac{1}{2} \times 600\times 0.039^2) - (0.05 \times 9.8 \times 1.5 \times sin 39)$

kinetic energy of glider = -0.376 j