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A 40-kg crate is pulled across a rough, horizontal surface with a force of 350 N at an angle of 25 degrees above the horizontal. The coefficient of kinetic friction is 0.14.

What is the horizontal acceleration (in m/s2) of the crate?

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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.

User Chris Rouffer
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