To determine the angle at which the student pulls the package we can use the concept of the net force acting on the package.
First let's calculate the net force acting on the package. The net force is the difference between the applied force (55 N) and the force of kinetic friction (15 N):
Net force = Applied force - Force of kinetic friction
= 55 N - 15 N
= 40 N
Next we can 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:
Net force = mass × acceleration
Now we can rearrange the equation to solve for the mass:
mass = Net force / acceleration
= 40 N / 10.0 kg
= 4.0 kg
With the mass of the package determined we can now analyze the forces acting on it. These forces include the applied force the force of kinetic friction and the gravitational force.
Since the package is on a horizontal surface the gravitational force acting on it can be decomposed into two components: a vertical component (mg pointing downwards and a horizontal component (0 N pointing along the surface.
Considering the vertical component (mg where g is the acceleration due to gravity (approximately 9.8 m/s^2 we have:
Vertical component (mg) = mass × g
= 10.0 kg × 9.8 m/s^2
= 98 N
The horizontal component of the gravitational force is canceled out by the normal force from the surface so we don't need to consider it.
Now let's analyze the horizontal forces. We have the applied force (55 N) and the force of kinetic friction (15 N) acting horizontally. These forces can be decomposed into two components: a horizontal component (F_h pointing towards the right and a vertical component (F_v pointing upwards.
Since the magnitude of the acceleration of the package is 2.5 m/s^2 we can calculate the horizontal component of the applied force using the equation:
F_h (applied force) = mass × acceleration
= 10.0 kg × 2.5 m/s^2
= 25 N
Similarly the horizontal component of the force of kinetic friction is calculated as:
F_h (force of kinetic friction) = Force of kinetic friction
= 15 N
At equilibrium (when the package is not accelerating horizontally the sum of the horizontal components of the applied force and the force of kinetic friction is zero:
ΣF_h = F_h (applied force) - F_h (force of kinetic friction) = 0
Therefore:
25 N - 15 N = 0
Now that we have the horizontal component of the applied force and the horizontal component of the force of kinetic friction we can use trigonometry to find the angle at which the student pulls the package.
Using the definition of tangent we have:
tanθ = F_v / F_h
Substituting the values:
tanθ = 98 N / (25 N - 15 N)
tanθ = 98 N / 10 N
θ = atan(9.8) ≈ 82.87°
Therefore the student pulls the package at an angle of approximately 82.87° to the horizontal.