- #1

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- Homework Statement
- A 1 kg block situated on a rough incline is connected to a spring of spring constant 100

N m–1 as shown in Fig. 6.17. The block is released from rest with the spring in the

unstretched position. The block moves 10 cm down the incline before coming to rest.

Find the coefficient of friction between the block and the incline. Assume that the

spring has a negligible mass and the pulley is frictionless.

- Relevant Equations
- F=ma

work energy theorem

-1/2kx^2 - umcosΘx +umgsinΘx = 0

I have used the work energy theorem like all source have shown me an have arrived at the right answer

where work one by all the forces is the change in kinetic energy

-1/2kx^2 - umgcosΘx +mgsinΘx = 0 is the equation

which becomes

-1/2kx -umgcosΘ+ mgsinΘ = 0

where k= spring constant

u= coefficient of friction

m=mass x= extension of spring and theta is the angle of inclination

there is no change in KE so it is zero

My problem is that can't we find the same result ( find u) with Newtons second law

my attempt at this is

-kx -umgsinΘ+umgcosΘ =ma when the spring is stretched and the body is at rest

since the body is at rest i have assumed a to be zero ( it could be non zero, i don't know.)

that gives us -kx -umgcosΘ +umgsinΘ = 0

if you compare this equation with the equation in red above you notice a factor of 1/2 missing from the term kx which bothers me

I am surely doing something wrong while using Newtons second law but i don't know what it is

A little help would be appreciated

where work one by all the forces is the change in kinetic energy

-1/2kx^2 - umgcosΘx +mgsinΘx = 0 is the equation

which becomes

-1/2kx -umgcosΘ+ mgsinΘ = 0

where k= spring constant

u= coefficient of friction

m=mass x= extension of spring and theta is the angle of inclination

there is no change in KE so it is zero

My problem is that can't we find the same result ( find u) with Newtons second law

my attempt at this is

-kx -umgsinΘ+umgcosΘ =ma when the spring is stretched and the body is at rest

since the body is at rest i have assumed a to be zero ( it could be non zero, i don't know.)

that gives us -kx -umgcosΘ +umgsinΘ = 0

if you compare this equation with the equation in red above you notice a factor of 1/2 missing from the term kx which bothers me

I am surely doing something wrong while using Newtons second law but i don't know what it is

A little help would be appreciated