Hello!

Denote Nana's mass as `m_1` and Popo's mass as `m_2.` Denote the incline angle as `alpha.`

Draw a force diagram for Nana. There are 4 forces: its own weight `m_1*g,` normal force `N` , friction force `F_f` and traction force from Popo `F_T.` By Newton's Second Law the sum...

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Hello!

Denote Nana's mass as `m_1` and Popo's mass as `m_2.` Denote the incline angle as `alpha.`

Draw a force diagram for Nana. There are 4 forces: its own weight `m_1*g,` normal force `N` , friction force `F_f` and traction force from Popo `F_T.` By Newton's Second Law the sum of these forces is equal to `m_1*a,` where `a` is her acceleration.

Consider axes x and y as shown on the picture. No doubt that acceleration `a` is parallel to the x-axis. Project forces on x and y.

y: `N-m_1*g*cos(alpha)=0,`

x: `F_f-F_T-m_1*g*sin(alpha)=m_1*a`

`F_f=mu*N=mu*m_1*g*cos(alpha),` `F_T=m_2*g.`

So we obtain

`a=g*(mu*cos(alpha)-m_2/m_1-sin(alpha)).`

[Compute `m_2/m_1+sin(alpha) approx 1.976` , `cos(alpha) approx 0.766` ]

**A)** `mu_1=0.7,` `a_1=g*(0.7*0.766-1.976)=-g*1.440=-14.110 (m/s^2).`

(minus means to the left, i.e. downward)

**B)** `mu_2=2.6,` `a_1=g*(2.6*0.766-1.976)=g*0.0157=0.154(m/s^2).`

Now acceleration is upward, i.e. Nana brakes (slows) but very slowly.

**C)** The speed of Nana is `V_1-t*a_2,` it becomes zero at `t_1=V_1/a_2.`

The distance is `V_1*t_1-(a_2*t_1^2)/2=(V_1^2)/(2a_2)=25/0.308 approx 81(m).`

Hope there is enough space.

**D)** `mu_3=2.8` and Popo's acceleration is `a_3,` so `F_T=m_2*(g+a_3)`

(or `m_2` transforms to `m_2*(1+a_3/g)` ).

The maximum `a_3` is found from

`0=g*(mu_3*cos(alpha)-(m_2/m_1)*(1+a_3/g)-sin(alpha)),`

or

`(m_2/m_1)*(1+a_3/g)=mu_3*cos(alpha)-sin(alpha),`

or

`1+(a_3)/(g)=(m_1/m_2)*(mu_3*cos(alpha)-sin(alpha)),`

or

`a_3=g*[(m_1/m_2)*(mu_3*cos(alpha)-sin(alpha)) - 1] = `

`=g*[3/4*(2.8*0.766-0.643) - 1]=g*0.126=1.24 (m/s^2).`

Low acceleration!

================================

**I made a mistake when supposed that `F_(T)=m_2g.` This is incorrect.**

The correct way is to consider Popo and forces acting on him.

One of them is `m_2g` downwards and another is `F_T` upwards (the same magnitude that pulls Nana downhill). And the net force is `m_2a` (Nana and Popo have the same magnitude of the acceleration). Popo's acceleration `a` is directed downwards, so to preserve the sign of a, we have to use vertical y-axis for Popo:

`F_(T)-m_2g=m_2a,`

or

`F_(T)=m_2(g+a).`

Now consider again

`mu*m_1g*cos(alpha)-F_(T)-m_1g*sin(alpha)=m_1a`

and substitute `F_(T)=m_2(g+a)` into it:

`mu*m_1g*cos(alpha)-m_2g-m_2a-m_1g*sin(alpha)=m_1a.`

From this equation

`a(m_1+m_2)=g(mu*m_1*cos(alpha)-m_1*sin(alpha)-m_2)`

and

`a=g*(mu*m_1*cos(alpha)-m_1*sin(alpha)-m_2)/(m_1+m_2).`

Now for the answers.

**A)** `mu_1=0.7,` `a_1=9.8*(0.7*60*0.766-60*0.643-80)/140 approx -6.05 (m/s^2)`

(minus means to the left, i.e. downward).

**B**) `mu_2=2.6, a_2=9.8*(2.6*60*0.766-60*0.643-80)/140 approx 0.064 (m/s^2).`

Even slower than I computed first...

**C**) the distance is `25/(2a_2)=25/0.128=195.3 (m).`

**D)** Was correct above.