Answer: 6.58 m/s^2 (approx).
Step-by-step explanation:
To find the horizontal acceleration of the crate, we need to consider the forces acting on it. The force of 350 N applied at an angle of 25 degrees above the horizontal can be broken down into its horizontal and vertical components.
The horizontal component of the applied force is given by:
F_horizontal = F_applied * cos(angle)
F_horizontal = 350 N * cos(25°)
F_horizontal = 350 N * 0.9063 ∴ cos 25° = 0.9063.
F_horizontal = 317.19 N
The frictional force acting against the motion of the crate is given by:
F_friction = coefficient_friction * normal_force
The normal force is equal to the weight of the crate, which can be calculated by the formula which is listed below:
normal_force = mass * gravity
normal_force = 40 kg * 9.8 m/s^2
normal_force = 392 N
Now, we can calculate the frictional force:
F_friction = 0.14 * 392 N
F_friction = 54.88 N
Since the crate is being pulled with a force greater than the frictional force, the net force in the horizontal direction is the difference between the applied force and the frictional force:
net_force_horizontal = F_horizontal - F_friction
net_force_horizontal = 317.19 N - 54.88 N
net_force_horizontal = 262.31 N
Finally, we can calculate the acceleration using Newton's second law:
net_force_horizontal = mass * acceleration
262.31 N = 40 kg * acceleration
acceleration = 262.31 N / 40 kg
acceleration = 6.58 m/s^2 (approx)
Therefore, the horizontal acceleration of the crate is approximately 6.58 m/s^2.