Question

A small 0.80-kg box is launched from rest by a horizontal spring as shown in the figure below. The block slides on a track down a hill and comes to rest at a distance d from the base of the hill. The coefficient of kinetic friction between the box and the track is 0.40 along the entire track. The spring has a spring constant of 37.0 N/m and is compressed 30.0 cm with the box attached. The block remains on the track at all times. (a) What would you include in the system? Explain your choice. spring, gravity, friction kinetic energy :) Score: 0 out of 0 Comment: (b) Calculate d. 1.03 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. m (c) Compare your answer with a similar situation where the inclined part of the track is frictionless. the same? Just no friction for inclined plane so you can't use TE1=TE2?

30.0 cm 50.0 cm 40

Answer 1

30cm oood 751 som 7.665 N-m for AB plane . work done - I kad² OR X37X(0:3)? friction = ung for = 0.4 X 0.889.8 fr = 3.136 N So workdone on friction surface 2.0 i frX0.3 0.9408 Nm. work done by spring to SD - 0.9408 Flejesept = 0.7242N-m o for baraller in inclind blan Net force 2. masing-uN on singsino ungeasoning 20.8X9.8 Sin40-0.4x0.8 c0540x918 5 .039 – 2.402 22.6370 work done = f. xo iss AN asing H = 2.637 X0.5 - 1.3185 N-m Page No

Date for brane CA Total work on above plano 0.7242 Q21982.+ 1.3185 W = Bengo 2.6427 work a forex disolacomont

umgd - 2.0927 0.4x0.8x9.8 x d = 2.0427 id= 2.0427 3.136 da 0.6513 Id= 665.13cm If surface is friction Less retardation is zero. So JH than Page 201