A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the spring, compressing it 0.20 m from its equilibrium position. The block is then released. If the block is not attached to the spring, how far up the incline will it travel before it stops?

Explanation:

The given data is as follows.

   Spring constant (k) = 78 N/m,    

 Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

                mgh sin

       h =

          =

          = 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

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explanation:

prolly a beacuse the momentuem the car has increases while going down hill