A block with mass m = 1.95 kg is placed against a spring on a frictionless incli

Question

A block with mass m = 1.95 kg is placed against a spring on a frictionless incline with angle theta = 30.0 degree. (The block is not attached to the spring.) The spring, with spring constant k = 18.4 N/cm, is compressed 23.0 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 point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

Explanation / Answer

Here ,

m = 1.95 Kg

theta = 30 degree

k = 18.4 N/cm = 1840 N/m

x = 23 cm

a) elastic potential energy in the spring = 0.50 * k * x^2

elastic potential energy in the spring = 0.50 * 1840 * 0.23^2

elastic potential energy in the spring = 48.7 J

b)

change in gravitational potential energy = 48.7 J

c)

let the distance is x

m * g * x* sin(theta) = 48.7

1.95 * 9.8 * x * sin(30 degree) = 48.7

solving for x

x = 5.1 m

the distance along the incline is 5.1 m

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