A block with mass m = 2.15 kg is placed against a spring on a frictionless incli

Question

A block with mass m = 2.15 kg is placed against a spring on a frictionless incline with angle theta = 34.5 degree (see the figure). (The block is not attached to the spring.) The spring, with spring constant k = 26 N/cm, is compressed 15.7 cm and then released, (a) What is the elastic potential energy of the compressed spring? (b) 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? (c) How far along the Incline is the highest point from the release point? Number _______________ Units Number ______________ Units Number ______________ Units

Explanation / Answer

a)elastic potential energy in the spring=1/2 k x^2

= 1/2 *26* 15.7^2 *10^-4

= 0.3204 Joule.

b)The elastic potential energy will convert into gravitational potential energy of spring when it is released.

so incraese in the gravitational potential energy = 0.3204 Joule.

c) 1/2 k x^2 = mgh h is height that block raised.... let " l "is diastance on inclined plane that the block moved up.

we have sin(theeta) = h/l

1/2 k x^2 = mglsin(theeta)

l= k x^2/ (2mgsin(theeta))

l= 2.68 cm

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