Two Blocks on two inclines(Friction)

In summary, a student has two ramps, both at an angle of 30o, with one being frictionless and the other having friction. When pushed with the same initial velocity, the block on the frictionless ramp travels 0.625 times farther than the block on the ramp with friction. To find the coefficient of sliding friction between the block and ramp with friction, we can use the formula Vf^2=Vi^2+2ad with the accelerations a1 = -g*sin(30) and a2 = -g*sin(30) - μg*cos(30). Setting the equations equal to each other and solving for μ, we get 2(-g)sin30(d)=2((-g)sin
  • #1
ryan1180
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0

Homework Statement



A student has two ramps, both of which are at an angle of 30o. Ramp 1 is frictionless and ramp 2 has friction. The student also has two blocks, one for each ramp. She pushes the blocks up the ramps with the same initial velocity. The block on ramp 2 only travels a fraction f = 0.625 as far as the block on ramp 1 before coming to a stop (i.e. d2 = 0.625*d1) .

Find the coefficient of sliding friction between the block and ramp 2.

[PLAIN]https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys100/fall10/homework/07/IE_friction_fraction/frictionfraction.gif



Homework Equations


Ff=muFn
Vf^2=Vi^2+2ad



The Attempt at a Solution



I know that
Vf^2=Vi^2+2ad

and my two formulas for acceleration are
a1 = -g*sin(30)
a2 = -g*sin(30) - μg*cos(30)

since Vi=0
Vf^2=2ad
since the Vf will be the same, the equations can be set equal to each other

2(-g)sin30(d)=2(-g)sin(30)-μgcos(30)(.625d)

I have tried this problem so many times and arrive at a final answer involving two variables. any help for finishing the problem would be greatly appreciated.
 
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  • #2
Are my formulas correct at least?
 
  • #3
ryan1180 said:
and my two formulas for acceleration are
a1 = -g*sin(30)
a2 = -g*sin(30) - μg*cos(30)

The accelerations are correct.

ryan1180 said:
since Vi=0
Vf^2=2ad

The final velocities are zero. The initial velocities are equal.

ryan1180 said:
the equations can be set equal to each other

2(-g)sin30(d)=2(-g)sin(30)-μgcos(30)(.625d)

You forgot the parentheses at the right-hand side around the acceleration. The correct equation is

2(-g)sin30(d)=2((-g)sin(30)-μgcos(30))(.625d)

ehild
 

Related to Two Blocks on two inclines(Friction)

1. What is the purpose of studying "Two Blocks on two inclines(Friction)"?

Studying the motion of two blocks on two inclines with friction allows scientists to better understand the effects of friction on objects in motion, as well as the relationship between forces and motion in a real-world scenario.

2. How does friction affect the motion of the blocks?

Friction acts as a resistive force that opposes the motion of the blocks, causing them to slow down and eventually come to a stop. This is due to the interaction between the surfaces of the blocks and the inclines, which results in a force that acts in the opposite direction of the motion.

3. What factors influence the amount of friction involved in this scenario?

The amount of friction between the blocks and the inclines depends on several factors, including the weight and surface texture of the blocks, the angle of the inclines, and the type of surface the blocks are sliding on.

4. How does the angle of the inclines affect the motion of the blocks?

The steeper the incline, the greater the force of friction acting on the blocks, resulting in a slower and more gradual descent. This is because a steeper incline means a greater component of the weight of the blocks is acting in the direction of the incline, increasing the friction force.

5. How does the presence of friction impact the acceleration of the blocks?

In the absence of friction, the blocks would accelerate down the inclines at a constant rate. However, the presence of friction reduces the net force acting on the blocks, leading to a slower acceleration and a longer amount of time for the blocks to reach the bottom of the inclines. This is because friction acts in the opposite direction of the motion, reducing the overall force and therefore the acceleration.

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