A box has a mass of 0.215 kg. It is on a very smooth (frictionless) horizontal surface and is acted upon by a force of 2.27 N at an angle of θ = 37.6° with the horizontal. What is the acceleration of the box?
Is the force below the normal, or above/
the force is above the normal
To find the acceleration of the box, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
In this case, the net force acting on the box is given by the force applied at an angle. However, since the surface is frictionless, there is no horizontal component of the force. Thus, we only need to consider the vertical component of the force.
The vertical component of the force can be calculated as:
F_vert = F * sin(θ),
where F is the given force (2.27 N) and θ is the angle with the horizontal (37.6°). Plugging in these values:
F_vert = 2.27 N * sin(37.6°).
Next, we can calculate the acceleration using Newton's second law:
F_net = m * a,
where F_net is the net force, m is the mass of the box (0.215 kg), and a is the acceleration. Since there is only a vertical force, the net force is the vertical component of the force:
F_net = F_vert.
Substituting the values into the equation:
2.27 N * sin(37.6°) = 0.215 kg * a.
Now, we can solve for the acceleration:
a = (2.27 N * sin(37.6°)) / 0.215 kg.
Using a calculator:
a ≈ 3.97 m/s².
Therefore, the acceleration of the box is approximately 3.97 m/s².