Calculating Toboggan Acceleration and Tension with Newton's Laws

[whatwhat11]

Two toboggans are connected by a rope. The first toboggan has a mas of 60 kg and the second has a mass of 40 kg. The two toboggans are pulled by a rope connected to the first toboggan. If the force exerted on the rope is 250N [E], and friction between the toboggans and the ice and be ignored, find:
(a) the acceleration of the toboggans.
(b) the tension in the rope connecting the toboggans

I found the acceleration to be 2.5m/s^2 and the tension in the rope to be 100N

Question 2 it asks, suppose a third toboggan of mas 25kg is connected tot he second toboggan in question 1. Find:
(a) the acceleration of the system of three toboggans.
(b) the force exerted by toggan one on toboggan two.
(c) the force exerted by the second toboggan on the last one.

I got 2m/s^2 for (a), 130N[E] for b and 50N[E] for c

But what I don't get is when it asks me to repeat question 2 with the coefficient of kinetic friction being 0.10.

I only found the acceleration being 1.02m/s^2. How can I find (b) and (c)?

 

[Confused too]

I know I'm not helping but how did you find 2b?

 

[whatwhat11]

I know I'm not helping but how did you find 2b?

i added the masses of the 40kg and 25kg toboggan then found FNet.
Fnet =ma
Fnet = (65)(2)
Fnet= 130N

 

[tiny-tim]

welcome to pf!

hi whatwhat11! welcome to pf!

(try using the X² button just above the Reply box )

I only found the acceleration being 1.02m/s^2. How can I find (b) and (c)?

same method

start with (c) …

you know the acceleration a, you know the friction force F, so the tension T is … ?

 

[whatwhat11]

hi whatwhat11! welcome to pf!

(try using the X² button just above the Reply box )


same method

start with (c) …

you know the acceleration a, you know the friction force F, so the tension T is … ?

The acceleration is 1.02 and the friction force for all of the toboggans is 122.5N. The answer key says that (b) has to equal 130N again and (c) has to equal 50N again.

For (c) I did Fg=mg
Fg = (25)(-9.8)
Fg = -245N

so the Ff=(0.10)(245)
Ff= 24.5N I am not sure what to do next ;/

 

[tiny-tim]

… so the Ff=(0.10)(245)
Ff= 24.5N

ok, so T - 24.5 = mass times acceleration

 

[whatwhat11]

hi whatwhat11! welcome to pf!

(try using the X² button just above the Reply box )same method

start with (c) …

you know the acceleration a, you know the friction force F, so the tension T is … ?

ok, so T - 24.5 = mass times acceleration

So (1.02)(25) = 25.5

-24.5 - 25.5 = -50N Right?

 

[tiny-tim]

So (1.02)(25) = 25.5

-24.5 - 25.5 = -50N Right?

(most people would have made those all pluses , but …)

yes

 

[whatwhat11]

Thank you :D