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A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a distance d_1 down the slope and then covers the distance d_2 along the horizontal terrain before stopping. Find the coefficient of kinetic friction mu_k between the sled and the ground, assuming that it is constant throughout the trip.
The coefficient is:{ d_1sin (theta)}/{d_2 + d_1cos(theta)}
To assess whether your calculations are reasonable, we'll analyze a similar situation to the one described in the problem introduction.
Suppose the same sled is released from the same height on the same slope. This time, however, assume that the coefficient of kinetic friction between the ground and the sled is a known quantity, mu, and, as before, constant throughout the trip. After the sled is released, it slides the same distance d_1 down the slope and then moves a certain (unknown) distance along the horizontal terrain before stopping. Find the distance d traveled by the sled from the end of the slope until it comes to a stop.
Express your answer in terms of the variables d_1, mu, and theta

Respuesta :

jolis1796

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

Distance 1:

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   (I)

 

Distance 2:

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     (II)

If we apply (I) + (II)

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ    (III)

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation III) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

shubhamchouhanvrVT

(1) The coefficient of kinetic friction will be given as  

[tex]\mu_k=\dfrac{Sin\Theta d_1}{d_2-cos\Theta d_1}[/tex]

(2) The distance d traveled by the sled from the end of the slope until it comes to a stop  

.[tex]d_2=\dfrac{d_1sin\theta}{\mu_k}[/tex]

What will be the coefficient of kinetic friction and the distance of the sled?

First, we will apply the Work energy theorem  

[tex]W=\Delta E[/tex]

[tex]W=-F_f\times d[/tex]

[tex]F_f=\mu_k\times N=\mu_kmg[/tex]

First, we find the distance one

[tex]-F_f\times d_1=E_f-E_i[/tex]

[tex]-(\mu_kmgcos\theta)\times d_1=(k_f+u_f)-(k_i+u_i)=(k_f+0)-(0+u_i)=K_f-u_i[/tex]

[tex]u_i=mgh=mgd_1sin\theta[/tex]

then

[tex]-(\mu_k mgcos\theta )\times d_1 =0.5v^2 -gd_1sin\theta[/tex]

[tex]-\mu_k gcos\theta d_1=0.5v^2 -gd_1sin\theta[/tex]............................(1)

Now for calculating the second distance

[tex]-F_f\times d_2=E_f-E_i[/tex]

[tex]-(\mu_k mg)d_2=(0+0)-(k_i+0)=-k_i[/tex]

[tex]k_i=0.5mv_i^2=0.5mv^2[/tex]

So  

[tex]-(\mu_k mg)d_2=0.5v^2[/tex]

[tex]\mu_k gd_2=0.5v^2[/tex]..............................(2)

Now add both the equation we will get

[tex]\mu_k g(d_2-cos\theta d_1)=v^2 -gd_1sin\theta[/tex]  ..........................(3)

Applying the equation (for the distance 1) we get v:

[tex]V_f^2=V_i^2+2ad=0^2(gsin\theta)d_1[/tex]

[tex]V-f^2=2gsin\thetad_1=V_i^2[/tex]    

Now from the  equation (3) we get

[tex]\mu_kg(d_2-cos\thetad_1)=2gsin\theta d_1-gd_1sin\theta[/tex]

[tex]\mu_k (d_2-cos\theta d_1)=sin\theta d_1[/tex]

[tex]\mu_k=\dfrac{sin\theta d_1 }{d_2-cos\theta d_1}[/tex]

Now to find the value of distance

We know the value of [tex]\mu_k[/tex]

By applying the work-energy theorem again

[tex]W=\Delta K[/tex]    ⇒  [tex]-F_fd_2=k_f-k_i[/tex]  

[tex]F_f=\mu_kmg[/tex]

[tex]k_f=0[/tex]

[tex]k_i=0.5mv^2=0.5m\times 2\times g\times sin\theta\times d_1=mgsin\theta md_1[/tex]

[tex]-\mu_k mgd_2=0-mgsin\thetad_1[/tex]

[tex]d_2=\dfrac{sin\thetad_1}{\mu_k}[/tex]

Thus

(1) The coefficient of kinetic friction will be given as  

[tex]\mu_k=\dfrac{Sin\Theta d_1}{d_2-cos\Theta d_1}[/tex]

(2) The distance d traveled by the sled from the end of the slope until it comes to a stop  

.[tex]d_2=\dfrac{d_1sin\theta}{\mu_k}[/tex]

To know more about the work-energy theorem follow

https://brainly.com/question/22236101

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