Kinetic Friction Homework: Solve for Acceleration of 880 N Crate

[unoriginal]

Homework Statement

A 880 N crate is being pulled across a level floor by a force F of 395 N at an angle of 25° above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.25. Find the magnitude of the acceleration of the crate.

Homework Equations

not quite sure?? was absent

The Attempt at a Solution

dont know where to begin

 

[unoriginal]

wow nevermind solved it...

found Fn first using the N of the crate and sin of the applied force, then i plugged that into

ma=uFn and found m by dividing 880 by 9.8 then just solved for a

 

[nlightner]

I would suggest starting by breaking down the problem into smaller components and identifying any relevant equations that can be used to solve for the acceleration. In this case, we are dealing with a crate being pulled at an angle with a known force and coefficient of kinetic friction.

One possible approach would be to use Newton's Second Law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the horizontal component of the pulling force minus the force of friction:

Fnet = Fpull - Ffriction

We can calculate the horizontal component of the pulling force by using trigonometry: Fpull = 395 N * cos(25°) = 355.54 N.

The force of friction can be calculated using the coefficient of kinetic friction and the normal force, which is equal to the weight of the crate (880 N) in this case:

Ffriction = μ * N = 0.25 * 880 N = 220 N

Plugging these values into the equation for net force, we get:

Fnet = 355.54 N - 220 N = 135.54 N

Now, we can use Newton's Second Law to solve for the acceleration:

Fnet = ma

135.54 N = (880 N) * a

a = 135.54 N / 880 N = 0.154 m/s^2

Therefore, the magnitude of the acceleration of the crate is 0.154 m/s^2.