Question

A block with mass m = 1.86 kg is placed against a spring on a frictionless incline with angle θ = 33.9° (see the figure). (The block is not attached to the spring.) The spring, with spring constant k = 25 N/cm, is compressed 28.1 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?

Chapter 08, Problem 031 A block with mass m = 1.86 kg is placed against a spring on a frictionless incline with angle 9 = 33.90 (see the figure). (The block is not attached to the spring.) The spring, with spring constant k = 25 N/cm, is compressed 28.1 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? k e (a) Number Units (b) Number Units (c) Number Units

Answer 1

Solution K Cinen mass of blode m=1186mg spring constant (2= 25 N/com 0 - 33.9° 0 = 33.99 = 2800lt/m a) potential energy of spring it ginen by x 2 500 x12811810-23? Pl= 98.70 5 PE= - ku2 b) since the inclined surface is friction less So, the energy will be conserved. Increase in P.E. = Change in cravitational Potential Energy P.EDU ->TAU= 98.70J

PAGE No. DATE c) sincer Gravitational P.f.a spring Pit a) mgh I kn 2 mg (h singgg - 98.70 => 1.86 x (98) L (sin 33.go) = 98.70 11 - 9.721