Pulling an object at an angle with friction -- find the Kinetic Energy

[omgbeandip]

Homework Statement

A boy pulls a sled (mass m=59.2 kg) along the ground using a force of magnitude F = 586 N inclined at an angle of θ=37.3° to the horizontal. There is friction between the sled and the ground with a coefficient of µk = 0.205. If the sled starts at rest, how much kinetic energy does the sled have after it is pulled a distance d =8.68 metres? Express your answer in Joules.

$m_{sled} = 59.2kg$
$F = 586N$
$\theta = 37.3°$
$\mu_k = 0.205$
$d = 8.68m$

Homework Equations

$W= \frac 1 2(m_{sled})(v)^2$
$\Sigma_x = F(\cos\theta) - \mu_kN = ma_x$
$V_f^2 = V_i^2 + 2a_xd$
$E_k = \frac 1 2 (m_{sled})(v)^2$

The Attempt at a Solution

Okay, so I have constructed a free body diagram to orient my question better. To find the acceleration in the problem, I have rearranged the $\Sigma_x = F(\cos\theta) - \mu_kN = ma_x$ to isolate a, which I determined to be $5.86_{m/s^2}$ . From that point on, I used kinematics formula $V_f^2 = V_i^2 + 2a_xd$ to find my $v_f$ , which was 10.09. Lastly, I used the $E_k = \frac 1 2 (m_{sled})(v)^2$ formula to find my final answer.

 

[berkeman]

Pulling the sled with a rope at an angle does two things. It accelerates the mass to the right according to the sum of all Fx forces. It also lightens the load some, since the vertical rope force on the sled pulls it up a bit.

So start with how much the sled is lightened by the vertical rope force, and THEN write the sum of the horizontal forces to get the horizontal acceleration. If the sled starts from rest, the acceleration will help you get the velocity you need to figure out the KE at the point you are asked for it. Makes sense?

 

[berkeman]

BTW, it's easier for us to read your work, and for us to quote it in our replies if you type your work into the forum window. You can either use clear text for simple equations, or you can start to learn and use LaTeX by looking at the tutorial under INFO, Help/How-To at the top of the page.

 

[omgbeandip]

Pulling the sled with a rope at an angle does two things. It accelerates the mass to the right according to the sum of all Fx forces. It also lightens the load some, since the vertical rope force on the sled pulls it up a bit.

So start with how much the sled is lightened by the vertical rope force, and THEN write the sum of the horizontal forces to get the horizontal acceleration. If the sled starts from rest, the acceleration will help you get the velocity you need to figure out the KE at the point you are asked for it. Makes sense?

Okay so I have rewritten my question using LaTeX (I think I'm using it properly). I realized that the sum of forces in the x direction equal $(m)(a)$ . Is there anything going on in the y-component for acceleration? I believe that I only have to use the x component because the sled isn't floating or sinking.

 

[berkeman]

Thanks for using LaTeX! So much easier to read your work!

W=12(msled)(v)2 W= \frac 1 2(m_{sled})(v)^2
Σx=F(cosθ)−μkN=max \Sigma_x = F(\cos\theta) - \mu_kN = ma_x
V2f=V2i+2axd V_f^2 = V_i^2 + 2a_xd
Ek=12(msled)(v)2

The Work is not equal to the final KE, since the process is lossy with energy going into the work done by friction.

Is there anything going on in the y-component for acceleration? I believe that I only have to use the x component because the sled isn't floating or sinking.

Yes, the vertical force of the rope decreases the normal force "N" that is used to determine the horizontal force due to friction on the sled. Can you factor that part in?

 

[omgbeandip]

So would the total work formula be: $$ W_{total} = \Delta_{Ek} - W_{friction} $$

And from previous notes of mine I have that $$ W_{friction} = -\mu(m)(g)(d) $$

Sorry for asking so many questions, I'm really struggling to find questions online that mimic this problem.

 

[haruspex]

And from previous notes of mine I have that $$ W_{friction} = -\mu(m)(g)(d) $$

Standard equations are useless if you do not know the context in which they apply. The equation you quote only applies in restricted situations. Better to scrub it from your notes and replace it with the most general case:
$W_{friction}=\int\mu_k\vec {F_{normal}}.\vec{ds}$. Or, dispensing with the vectors, and understanding that the displacement must be parallel to the friction force, $W_{friction}=\int\mu_kF_{normal}.ds$.
Generally the coefficient is constant, allowing the further simplification $W_{friction}=\mu_k\int F_{normal}.ds$.
If the normal force is constant too then we can get rid of the integral altogether: $W_{friction}=\mu_k F_{normal}\Delta s$.
But what is F_normal? Often it is just mg, but also quite often it is not. What is it in this question?