Question

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 46.0 ∘ above the horizontal. The glider has mass 9.00×10−2 kg . The spring has 670 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.80 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. the distance of the spring was originally compressed is x = 5.84×10−2 What is the kinetic energy of the glider at this 0.8m?

Answer 1

Here

$W_{total}$  = 1/2 m $v^{2}$ - 0 = final kinetic energy - initial kinetic energy = final kinetic energy, as initial kinetic energy = 0

here we have to understand that Work done by spring is counteracted by the work done by gravity.

$\therefore$   $W_{total}$ =

= mg L cos$\theta$ where m = mass of the glider = 9.00×10−2 kg

g = Acceleration due to gravity = 9.8 m / $s^{2}$

L = maximum distance travelled by the glider in contact with the spring= 0.8m

=  9.00×10−2 $\times$ 9.8 $\times$ 0.8 $\times$  cos ( 90-46)

= 70.56 $\times$ 10−2 $\times$ cos 54

= 7056 $\times$$10^{-4}$$\times$ 0.58

= 4092.48 $\times$$10^{-4}$ J

     where K = spring constant = 670 N/m

and the distance of the spring originally compressed is x = 5.84×10−2 m

= 1 / 2 $\times$ 670 $\times$ 5.84× 5.84×$10^{-4}$ = 335 $\times$ 5.84× 5.84×$10^{-4}$

= 11425.76 ×$10^{-4}$ J

$\therefore$   $W_{total}$ =

= 11425.76 ×$10^{-4}$ -   4092.48 $\times$$10^{-4}$ = 7333.28 $\times$$10^{-4}$ J

$\therefore$  the kinetic energy of the glider = 7333.28 $\times$$10^{-4}$ J