What is the speed of a block sliding down an inclined plane with friction?

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The discussion focuses on calculating the speed of a 15.0 kg block sliding down a 17.5-degree inclined plane with a friction coefficient of 0.085 over a distance of 4.0 meters, starting from rest. The calculations involve determining the forces acting on the block, including gravitational and frictional forces, and applying Newton's second law to find acceleration. The final speed calculated is approximately 4.15 m/s. An alternative method suggested for solving the problem is using energy equations. The calculations and approach appear to be correct, with minor noted typos.
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Homework Statement



A board is inclined at an angle of 17.5 degrees. The coefficient of friction between a 15.0 kilogram block and the board is .085. what will be the speed of the block if it slides down the board a distance of 4.0 meters, starting from rest?
 
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Hi sskicker23,

sskicker23 said:

Homework Statement



A board is inclined at an angle of 17.5 degrees. The coefficient of friction between a 15.0 kilogram block and the board is .085. what will be the speed of the block if it slides down the board a distance of 4.0 meters, starting from rest?


What have you tried so far?
 
what have it tried so far:

Friction
mu* Fnormal
.085*15*9.8*Cos(17.5)=11.91669

Fnormal
M*G*Cos(17.5)= 140.1964N

Fperp
M*G*Cos(17.5)= 140.1964N

Fgravitational
m*g
15*9.8=147N

Fparallel= m*g*sin(17.5) =140.1964N

axmax= -gsin(17.5)+Mu*G* Cos(17.5)
A=.9.8*sin(17.5)+9.80*Cos(17.5)
A= 2.1525

v^2=Vi^2+2AX
V^2=0+2*2.1525*4
v=4.15m/s
 
sskicker23 said:
what have it tried so far:

Friction
mu* Fnormal
.085*15*9.8*Cos(17.5)=11.91669

Fnormal
M*G*Cos(17.5)= 140.1964N

Fperp
M*G*Cos(17.5)= 140.1964N

Fgravitational
m*g
15*9.8=147N

Fparallel= m*g*sin(17.5) =140.1964N

axmax= -gsin(17.5)+Mu*G* Cos(17.5)
A=.9.8*sin(17.5)+9.80*Cos(17.5)

This line has a few problems, but I think they are just typos.

A= 2.1525

v^2=Vi^2+2AX
V^2=0+2*2.1525*4
v=4.15m/s

That looks like the right answer to me.


(An alternative approach to this problem would be to use the energy equation.)
 
Thank you for your help!
 

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