Find The Distance And Speed By Using The Work And Energy Theorem

[cbarker1]

Dear Everybody,

I need some help to help the distance for part a. b) If the block comes back down after it was push, what is the speed back down.

A 3.25 kg block starts with a speed of 15 m/s at the bottom of a plane inclined at 35° to the horizontal. The coefficient of sliding friction between the block and plane is μ_k = .30.

thanks
Cbarker1

 

[MarkFL]

As the block slides up the incline plane, what forces are working to retard the motion of the block?

 

[cbarker1]

As the block slides up the incline plane, what forces are working to retard the motion of the block?

the force that move block up the incline plane and the friction force as well as gravity and normal forces.

 

[MarkFL]

the force that move block up the incline plane and the friction force as well as gravity and normal forces.

Well, the block moves up the incline because of its initial momentum, but as it slides, there is the force of friction and the force of gravity working to retard its motion. Can you quantify these forces?

 

[cbarker1]

$$f=\mu_k*N$$ and
$mg=<mgsin\theta,mgcos\theta>$

 

[MarkFL]

The magnitude of the force of kinetic friction is:

\(\displaystyle f_k=\mu_kn=u_kmg\cos(\theta)\)

And the magnitude of the force of gravity is:

\(\displaystyle f_g=mg\sin(\theta)\)

Now, as the block reaches it stopping point, at which time it has increase its height by $h$ and slid a distance of $x$ along the plane, we find:

\(\displaystyle \cos(\theta)=\frac{\sqrt{x^2-h^2}}{x}\)

\(\displaystyle \sin(\theta)=\frac{h}{x}\)

Now, what is the initial energy of the block? What amount of work is required to bring the block to rest? What work have the two forces done during the time it took for the block to come to rest? What energy does the block have when it has come to rest?

 

[cbarker1]

The magnitude of the force of kinetic friction is:

\(\displaystyle f_k=\mu_kn=u_kmg\cos(\theta)\)

And the magnitude of the force of gravity is:

\(\displaystyle f_g=mg\sin(\theta)\)

Now, as the block reaches it stopping point, at which time it has increase its height by $h$ and slid a distance of $x$ along the plane, we find:

\(\displaystyle \cos(\theta)=\frac{\sqrt{x^2-h^2}}{x}\)

\(\displaystyle \sin(\theta)=\frac{h}{x}\)

Now, what is the initial energy of the block? What amount of work is required to bring the block to rest? What work have the two forces done during the time it took for the block to come to rest? What energy does the block have when it has come to rest?

the initial of the block is zero. $W=F*d*cos \phi$ where $\phi$ is 0 or $ \pi$
We have to add those forces together

\(\displaystyle f_g=mg\sin(\theta)\)+
\(\displaystyle f_k=\mu_kn=u_kmg\cos(\theta)\)

 

[MarkFL]

Here's the way I would set this up. State the initial energy of the block (it has kinetic energy initially):

\(\displaystyle E_i=K_i=\frac{1}{2}mv_i^2\)

State the final energy of the block (it has gravitational potential energy):

\(\displaystyle E_f=P_f=mgh=mgx\sin(\theta)\)

Now, during the sliding of the block, some energy is lost to friction, and by the work-energy theorem, this is the work done by friction, hence:

\(\displaystyle \Delta E=-f_kx=-\mu_knx=-u_kmg\cos(\theta)x\)

\(\displaystyle E_f-E_i=-u_kmg\cos(\theta)x\)

\(\displaystyle mgx\sin(\theta)-\frac{1}{2}mv_i^2=-u_kmg\cos(\theta)x\)

Solve this for $x$...what do you get?

 

[cbarker1]

Here's the way I would set this up. State the initial energy of the block (it has kinetic energy initially):

\(\displaystyle E_i=K_i=\frac{1}{2}mv_i^2\)

State the final energy of the block (it has gravitational potential energy):

\(\displaystyle E_f=P_f=mgh=mgx\sin(\theta)\)

Now, during the sliding of the block, some energy is lost to friction, and by the work-energy theorem, this is the work done by friction, hence:

\(\displaystyle \Delta E=-f_kx=-\mu_knx=-u_kmg\cos(\theta)x\)

\(\displaystyle E_f-E_i=-u_kmg\cos(\theta)x\)

\(\displaystyle mgx\sin(\theta)-\frac{1}{2}mv_i^2=-u_kmg\cos(\theta)x\)

Solve this for $x$...what do you get?

$\frac{1}{2}mv_i^2)/(u_kmg\cos(\theta)+(mg\sin(\theta))=x$

 

[MarkFL]

$mgx\sin(\theta)-\frac{1}{2}mv_i^2=-u_kmg\cos(\theta)x$

$-(mgx\sin(\theta)-\frac{1}{2}mv_i^2)/u_kmg\cos(\theta)=x$

This is what I get:

\(\displaystyle x=\frac{v_i^2}{2g\left(\sin(\theta)+u_k\cos(\theta)\right)}\)

edit: THis appears to match your edited post, although I simplified. :)

Now, can you reverse the process to determine the speed of the block at the bottom of the plane, when it slides back down?

 

[cbarker1]

Do I use the value that I get to find the value of the speed of block back down?

 

[MarkFL]

Do I use the value that I get to find the value of the speed of block back down?

As the block slides back down the plane, it initially has gravitational potential energy, and then when it reaches the bottom (after sliding distance $x$), it will have kinetic energy (with speed $v_f$), and just as with the first part, there is energy lost due to friction.

See if you can set this up similarly to how I set up the first part...:)

 

[MarkFL]

To follow up:

State the initial and final energies:

\(\displaystyle E_i=mgx\sin(\theta)\)

\(\displaystyle E_f=\frac{1}{2}mv_f^2\)

State that the energy lost is due to the work done by the force of kinetic friction:

\(\displaystyle E_f-E_i=-u_kmg\cos(\theta)x\)

Plug in the known energies:

\(\displaystyle \frac{1}{2}mv_f^2-mgx\sin(\theta)=-u_kmg\cos(\theta)x\)

Solve for $v_f$:

\(\displaystyle v_f^2=2g\left(\sin(\theta)-u_k\cos(\theta)\right)x\)

Plug in the value we found for $x$ in the first part of the problem:

\(\displaystyle v_f^2=2g\left(\sin(\theta)-u_k\cos(\theta)\right)\left(\frac{v_i^2}{2g\left(\sin(\theta)+u_k\cos(\theta)\right)}\right)\)

Hence:

\(\displaystyle v_f=v_i\sqrt{\frac{\sin(\theta)-u_k\cos(\theta)}{\sin(\theta)+u_k\cos(\theta)}}\)