10 A block is released from rest and allowed to slide down a 36.1 o incline with

Question

10

A block is released from rest and allowed to slide down a 36.1o incline with a kinetic friction coefficient of 0.2. After sliding 99.3 cm down the incline, it slides across a frictionless horizontal surface and encounters a spring. The far end of the spring is attached to a wall (see figure for problem 105 on page 231). The spring compresses from the initial length of 15.0 cm down to the maximum compression of 11.0 cm.

After the compression, the spring rebounds and shoots the block back up the incline. How far along the incline does the block travel before coming to rest?

**********

CHEGG,

Please start with (show how you derive from) simple/basic formulas, and please clearly label and show ALL your work. Thank you!

Explanation / Answer

sin(36.1)=height of the incline/length of the incline

given that length of the incline is 99.3 cm (or 0.993 m)

==>height of the incline=0.993*sin(36.1)=0.58507 m

let mass of the block is M kg.

initial potential energy of the block=mass*g*height of the block=M*g*0.58507=M*9.8*0.58507=5.7337*M J

normal force on the incline=M*g*cos(36.1)=7.9183*M N

then kinetic friction force=friction coeficient*normal force=0.2*7.9183*M=1.5837*M N

work done against friction while travelling 0.993 m along the incline=1.5837*M*0.993=1.5726*M J

at the bottom of the incline, potential energy of the block=0

hence kinetic energy=initial potential energy-work done against friction=5.7337*M-1.5726*M=4.1611*M J

as the horizontal surface is frictionless, there wont be any energy less when the block travels towards the spring and when it rebounds.

hence the block will reach the bottom of the incline after rebounding with total eenrgy=4.1611*M J

now , if the block finally travels to a distance of d m, before coming to stop , height reached=d*sin(36.1)=0.5892*d m

work done against friction while moving up a distance of d=friction force*distance=1.5837*M*d J

final potential energy=mass*g*height=M*g*0.5892*d=5.7741*M*d J

using conservation of energy principle:

initial total energy-work done against friction=final total energy

==>4.1611*M-1.5837*M*d=5.7741*M*d

==>4.1611=(1.5837+5.7741)*d

==>d=4.1611/(1.5837+5.7741)=0.56554 m

hence the block will travel 56.554 cm along the incline before coming to rest.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.