What is the acceleration of the cart

To move a large crate across a rough floor, you push on it with a force F at an angle of 21 degrees, below the horizontal. Find the acceleration of the crate, given that the mass of the crate is m= 27 kg, the applied force is 333 N and the coefficient of kinetic friction between the crate and the floor is 0.53.

F = Ma

Since the crate is not accelerating vertically, the sum of the Vertical forces equal zero. There are three forces acting on the crate in the vertical direction: the weight of the crate (mg) in the downward direction, the normal force (n), and the vertical component of the 333N force (333N x sin21) in the downward direction(because the angle is below the horizon, the vertical component is directed downward). The normal force opposes the combined force of weight and the vertical component (333N x sin21).

n - mg - 333 x sin21 = 0 Solve for n

n = mg + 333 x sin21

n = (27kg x 9.8) + 119 = 384N

We can now solve for the horizontal acceleration. The horizontal component of the
333N force (333N x cos21) minus the frictional force (Uk x n) will give us the net force acting on the crate in the horizontal direction.

Fnet = 333N x cos21 - .53 x 384 = 107N

Now we can solve for acceleration.

F = ma

a = Force/mass = 107N/27kg = 4.0 meters per second squared

F=ma ---> a=F/m

F=Fx - Ffriction

Fx=333cos(21)= 310.882 N

Ffriction = Coefficient of kinetic friction multiplied by the Normal force of the box
= (0.53)(27 kg * 9.8 m/s^2) = 140.238 N

a= (Fx-Ffriction) / (27 kg) = (310.882 N - 140.238 N) / (27 kg) = 6.32 m/s^2

let $=22 degree
then Reaction force on the block will be mg-Psin$ where P=333 N =applied force.
So frictional force will be 0.53x(Psin$-mg)
So the equation of force will be P cos $ - 0.53(mg-Psin$)=mf where f = acceleration.

Or f= (333 cos22- 0.53(27x9.8-333sin22))/27=8.68 m/sec^2
So accleration will be 8.68 m/sec^2.