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

In summary, using the given information and equations, the speed of the 15.0 kilogram block sliding down the board at an angle of 17.5 degrees with a coefficient of friction of .085 and a distance of 4.0 meters is 4.15 m/s. An alternative approach using the energy equation is also possible.
  • #1
sskicker23
3
0

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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  • #2
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?
 
  • #3
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
 
  • #4
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.)
 
  • #5
Thank you for your help!
 

What is a sloped incline with friction?

A sloped incline with friction is a type of physics problem where an object is placed on an inclined plane and has to overcome the force of friction in order to move up or down the incline. This type of problem is commonly used to demonstrate concepts related to forces, motion, and energy.

How is the force of friction calculated on a sloped incline?

The force of friction on a sloped incline can be calculated using the formula: Ff = μN, where Ff is the force of friction, μ is the coefficient of friction, and N is the normal force. The normal force is equal to the weight of the object multiplied by the cosine of the angle of the incline.

What factors affect the force of friction on a sloped incline?

The force of friction on a sloped incline is affected by several factors including the weight of the object, the coefficient of friction between the object and the incline, and the angle of the incline. Rougher surfaces and heavier objects will typically have a higher force of friction.

How does the angle of the incline affect the motion of an object on a sloped incline?

The angle of the incline affects the motion of an object on a sloped incline by changing the normal force and the force of friction. As the angle of the incline increases, the normal force decreases and the force of friction increases, making it more difficult for the object to move up the incline.

What is the difference between kinetic and static friction on a sloped incline?

Kinetic friction is the force of friction that acts on an object while it is in motion on a sloped incline. Static friction is the force of friction that acts on an object when it is at rest on a sloped incline. The force of kinetic friction is usually lower than static friction because the object is already in motion and has overcome the initial force of static friction.

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