Calculating Mass Sliding on a Slope

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hodgepodge
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Homework Statement



A 15 kg mass is sliding at 30 m/s when it encounters a slope (smooth=no friction) and descends 60 m. It encounters a horizontal rough stretch (friction = 80N in opp. direction). Calculate the velocity at x=0 (rough stretch starts here), x=50, x=100, x=200. Calculate where it comes to a stop.

MASS
_______
h =60m \
...\friction = 80N for horizontal line only (indicated by ~)
....\~~~~~~~~~~
.....x=0 x= 50...
 
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i do not know how to do the sliding part to get the initial speed, but i think from x=0 i would use w=K final- K initial, which would be (friction x m x g)x= 1/2 x mass x final velocity -1/2 x mass x initial velocity

also, when i calculate the velocity for x=50, do i use the final velocity for x=0 for the initial velocity for x=50, or do i use the velocity from when it came down the slope for initial velocity for x=50
 
Last edited:
i guess energy would be conserved

this would be easrier but i have a lack of formulas
 
hodgepodge said:
i guess energy would be conserved

this would be easrier but i have a lack of formulas
That would be correct. Hopefully, the formulas will come to you as we talk though it.

What two forms of energy are involved in the block sliding down the slope?
 
kinetic and potential
 
hodgepodge said:
kinetic and potential
Good. So you know that the sum of the potential and kinetic energy at the top and the bottom of the slope must be equal.

What is the value of the total energy at the top of the slope?
 
Total energy = PE + KE
TE = mgh+1/2 x m x v squared
TE= 9000+6750
TE=15750

i used g=10
 
can anybody help with this?
 
As you know energy is conserved.

[tex]E_{total\ before}=E_{total\ after}[/tex]

[tex]Eg1+Ek1+Et=Eg2+Ek2[/tex]

[tex]mgh_{1}+\frac{1}{2}mv_{1}^2-F_{f} \triangle d = mgh_{2} +\frac{1}{2}mv_{2}^2[/tex]

always make the lowest point have a gravitational energy of zero
 
wat is Ff change in d?
 
Ff (80N) is the force in friction and Change in d is x
 
thank you so much bright wang