How Is Friction Affecting the Motion of a Block on a Horizontal Surface?

[DeltaIceman]

Homework Statement

A 3.52 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=12.6N at an angle theta=31.0degrees above the horizontal. The floor has a coefficient of kinetic friction µk = 0.04, what is the magnitude of the frictional force on the block when the block is moving? There is a second part which is What is the magnitude of the acceleration of the block when friction is being considered? Although I think I am forgetting forumals or something. If someone can just send me along the right track I think I can figure it out. Thanks!

m = 3.52
F= 12.6
Theta = 31 degrees
µk = 0.04

Homework Equations

Sine and cosine functions.
µkFn= fk

The Attempt at a Solution

 

[LowlyPion]

You mean you think you are missing F = ma?

(Less of course the correction you will need to make for the retarding force of friction.)

 

[thomate1]

To find the magnitude of the frictional force, we can use the formula µkFn = fk, where µk is the coefficient of kinetic friction and Fn is the normal force exerted by the floor on the block. We can find the normal force by using the cosine function to find the component of the tension force in the vertical direction, which is equal to the normal force.

Fn = Fcosθ = (12.6N)(cos31°) = 10.9N

Now, using the formula µkFn = fk, we can find the magnitude of the frictional force:

fk = (0.04)(10.9N) = 0.44N

For the second part, we can use Newton's second law, F = ma, to find the acceleration of the block when friction is considered. The total force acting on the block is the tension force in the horizontal direction, minus the frictional force:

Fnet = Ft - fk = (12.6N)(sin31°) - 0.44N = 5.96N

Using F = ma, we can solve for acceleration:

a = Fnet/m = (5.96N)/(3.52kg) = 1.69 m/s^2

Therefore, the magnitude of the acceleration of the block when friction is considered is 1.69 m/s^2.