A 2.00 kg block is placed against a spring on a frictionless 30.0° incline (Fig.

Question

A 2.00 kg block is placed against a spring on a frictionless 30.0° incline (Fig. 8-33). (The block is not attached to the spring.) The spring, whose spring constant is 19.6 N/cm, is compressed 18.0 cm and then released.

What is the elastic potential energy of the compressed spring (in Joules)?

What is the change in the gravitational potential energy of the block-Earth system as the block moves from the release point to its highest point on the incline (in Joules)?

How far along the incline is the highest point from the release point (in meters)?

Explanation / Answer

(a) PE of spring is given by (1/2) k x2    so

   PE = (1/2) * 1960 N/m * (0.18 m)2 =   31.75 Joules

(b) since energy is conserved, all of the energy that was stored in the spring now becomes gravitational potential energy. So the change in grav potential energy must be +31.75 Joules

(c) Two steps now... grav PE = mgh   so we use this to find the vertical height:

    h = PE / mg = 31.75 / 2.00 * 9.80 =   1.62 meters

and the distance along the incline is related to the height by simple trig:

     height = distance * sin30      or

distance along incline = height / sin30 = 1.62 / sin30 = 3.24 meters

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