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

[sskicker23]

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?

 

[alphysicist]

Hi sskicker23,

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?

 

[sskicker23]

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

 

[alphysicist]

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.)

 

[sskicker23]

Thank you for your help!