Help with Physics 11 Kinematics.

[aznpiggy]

Homework Statement

A 5.0kg concrete block accelerates down a 34° slope at 4.2m/s^2. Find the coefficient of friction between the block and the slope.

Homework Equations

Fg=5*9.8=45

x=45sin30

y=45cos30

The Attempt at a Solution

a=(mgsinθ-Ff)/m → Ff=mgsinθ-ma

Thanks

 

[TaxOnFear]

Welcome to PF,
Have you drawn your free body diagram? How did you get your x and y components? The question states that the slope is at a 34° angle to the horizontal, not 30°.

Hint: Friction force = μN

Hint 2: Remember to add units to EVERYTHING.

 

[aznpiggy]

Yes, I have drawn a FBD with x and y components that is perpendicular to the ramp.

I used the Fg to find the x by going Fgsin30°. Which would be the same as Fn.

Where would I go from this step?

 

[TaxOnFear]

I used the Fg to find the x by going Fgsin30°. Which would be the same as Fn.

Why are you using 30° as your angle when it states in the question that it is 34°?

Your x component isn't the same as your y component, which would be the normal force (Fn). You must also consider the acceleration of the concrete block in your calculations.

 

[asteriatic]

for reaching out for help with Physics 11 Kinematics. To find the coefficient of friction, we can use the equation Ff=μN, where Ff is the force of friction, μ is the coefficient of friction, and N is the normal force. In this case, the normal force is equal to the weight of the block, which is 45N. Using the equation Ff=mgsinθ-ma, we can substitute in the known values to solve for the coefficient of friction. We get Ff=45sin34°-5kg*4.2m/s^2=14.4N. Then, we can plug this value into Ff=μN to get μ=14.4N/45N=0.32. Therefore, the coefficient of friction between the block and the slope is 0.32. I hope this helps with your homework. Good luck!