Question

A small block lies against a rubber band at the bottom of an incline angled 20.0 degrees from the horizontal. The block has a mass of 0.05 kg. The rubber band has a spring constant of 99.0 N/m.

a. If you pull the block on the rubber band to stretch it 0.0508 m backwards from its equilibrium, how much elastic potential energy is in the rubber band-block-earth system?

b. After you let go, when the rubber band reaches equilibrium again, what is the gravitational potential energy of the rubber band-block-earth system (in J)?

c. If the incline surface did 0.002339 J of work through friction on the system while the rubber band pushed back to its equilibrium point, what is the kinetic energy of the rubber band-block-Earth system when the rubber band reaches equilibrium (in J)?

d. If the block displaces 0.9342m up the incline, what is the system's gravitational potential energy at that point (in J)?

e. How much work did the incline do on the system through friction during this displacement up the incline?

Answer 1

a) Elastic PE of rubber band=1/2 Kx^2 = 0.5 (99)(0.0508 )^2 =0.1277 J apprx

b) for this part I am assuming that there is no friction and entire elastic PE is used by block to gain the height

1/2kx^2( elastic PE ) = mgh ( Gravatational PE)

Therefore, gravational potential energy of the rubber band-block-earth system= 0.1277

c) Elastic PE = 1/2 mv^2 + work done to overcome friction

0.1277 = KE of block + 0.002339

KE of block =0.125361 J

d) height gained by block = 0.9342m (sin 20) = 0.9342m (0.342 ) =0.319 m apprx

system's gravitational potential energy at that point = ( 0.05)(9.8)( 0.319 ) =0.14377 J apprx