Object sliding with friction into a spring Find compression

[SamLing2000]

Homework Statement

A 1.18 kg block slides across a horizontal surface directly toward a massless spring with spring constant 5,803 N/m. The surface is frictionless except for a rough patch of length 0.42 m that has coefficient of kinetic friction 0.367. The initial velocity of the block is 2.49 m/s directed straight toward the spring. What is the maximum compression of the spring?

What i obtained form the question:
v0 = 2.49
friction coefficient 0.367
m=1.18
d=0.42
Spring constant = 5803

Homework Equations

W=Fd
W=Δk
k=1/2mv^2

The Attempt at a Solution

What I first did was find the knetic energy before the object hit the spring so i got
kf = .5(1.18)(vf)
ki = .5(1.18)(2.49)

i replaced vf with the equation to obtain equation is vf^2 = vi^2 + 2ax

kf = .5*(1.18)(2.49^2+2a(0.42))
Now i have W=Fd which i made W=(1.18)(a)(.42)

then i set the equations equal to each other

.5(1.18)(2.49^2+2a(.42))-.5(1.18)(2.49)=(1.18)(a)(.42)

solved for a = -1.029188469

used this a to find vf = 2.309887808

now i have the velocity it hits the spring with I then again brought back the work equation to find

W = kf-ki since final knetific force is 0 i have
W = -.5(1.18)vf^2 = Fd = ((-5803x)x)

then i solved for x as 0.0232911326
as demanded I entered 2.33 cm. But i was marked wrong. Can anyone help? That'll be greatly appreciated.

 

[flatmaster]

You are getting confused with your initial and final velocities. You need to break your problem into two separate problems.

I. Block slows down on friction section.

II. Block compresses spring.


The final velocity for section I will become the initial velocity for section II of the problem.

 

[SamLing2000]

That is what I did. If you follow my work.
Is the procedures right? I did i simply do my math wrong?

 

[flatmaster]

Ok, I think I found your mistake. In general, it's best to leave everything as variables and put numbers in at the end. otherwise, you end up with a jumble of numbers and no idea where they came from.

You weren't allowed to do this step.

".5(1.18)(2.49^2+2a(.42))-.5(1.18)(2.49)=(1.18)(a)(.42)"

The LHS is the kenitic energy after going through the rough patch. The RHS is the energy LOST to the friction.

Go back to vf^2 = vi^2 + 2ax. You will use this to find the velocity after the rough section. However, you still need to find the acceleration. Look up your equations for the force of friction. you will use this equation and F=ma to get the acceleration.

 

[SamLing2000]

Thanks for the response, but I am a bit confused as to what to do, and what you said.
First of all I'm curious as to why the RHS is the energy LOSt to friction.
Secondly, to find the acceleration I assume i must make 1.18(a)*.42 = uk(N) which is

1.18a(0.42)= 0.367(1.18)(9.8)
is that correct?
Once again thanks for taking the time to help me out.

 

[flatmaster]

You're close, but you multiplied by the distance for some reason.

We need to find the acceleration in the rough area. Start with Newton's Second law. We might as well solve for acceleration.

F=ma.

a = F/m

In this area, we know that the force is being caused by friction.

F = uN = umg

Where u is the coefficient of friction and N is the normal force.

Now plur this into the above equation.

a = umg/m

At this point, what can you do with the term for mass in the equation?

Notice that I didn't type any actual numbers. When working a problem, keep all your numbers as variables until the end.

 

[SamLing2000]

So I tried again using the a from that equation and I was still unfruitful. What I did was
found a = ug

vf = √(vi^2-2aΔx)

i used the minus because when i used the plus the velocity actually went up, and i assumed because it was the deceleration caused by the friction it should be a negative number for a.

and using that i have

x = √(.5 *m *vf^2 /5986)

the answer was still incorrect.

 

[azizlwl]

I prefer to solve this kind of problem by work-energy theorem rather than kinematic.
By work-energy theorem single equation can be formed to solve the problem.

ΔKE +ΔPE= Work done.

 

[Zubeen]

If you follow a simple law , the law of conservation of energy, you'll get the answer.
The initial energy of object is 1/2mv^2.
When it moves over distance "d" having friction force coefficient as "k", the friction force does the work and reduced its energy by same amount as work and is given by - mg*k*d.
Finally the block will compress the spring and will come to rest, but since energy must be conserved, the energy must remain there.
But actually what happens is that the energy gets stored in the spring as potential energy.
This energy (given by 1/2kx^2, k is stiffness constant of spring ) is equal to energy left out with block after it crossed the distance with friction.
Just do the mathematical calculations with these pictures in mind and you'll get the answer.
Zubeen

 

[MrWarlock616]

If you follow a simple law , the law of conservation of energy, you'll get the answer.
The initial energy of object is 1/2mv^2.
When it moves over distance "d" having friction force coefficient as "k", the friction force does the work and reduced its energy by same amount as work and is given by - mg*k*d.
Finally the block will compress the spring and will come to rest, but since energy must be conserved, the energy must remain there.
But actually what happens is that the energy gets stored in the spring as potential energy.
This energy (given by 1/2kx^2, k is stiffness constant of spring ) is equal to energy left out with block after it crossed the distance with friction.
Just do the mathematical calculations with these pictures in mind and you'll get the answer.
Zubeen

Yeah...I don't think there's any need to find acceleration.