A student moves a box of books down the hall

by pulling on a rope attached to the box. The
student pulls with a force of 179 N at an angle
of 29.8

above the horizontal. The box has a
mass of 42.2 kg, and µk between the box and
the floor is 0.2.
The acceleration of gravity is 9.81 m/s
2
.
Find the acceleration of the box.
Answer in units of m/s
2

remember that fricion is dependent on Normal force. But the vertical component of pulling is upward, so it lessens weight.

Net force=mass*a
horizonalpulling-friction=ma
solve for a.

To find the acceleration of the box, we need to determine the net force acting on it.

First, we need to resolve the pulling force into its horizontal and vertical components.

The horizontal component of the force can be found using the equation:

F_horizontal = F * cos(theta)

where F is the pulling force (179 N) and theta is the angle above the horizontal (29.8 degrees).

F_horizontal = 179 N * cos(29.8 degrees) = 155.02 N

Next, we need to calculate the frictional force. The frictional force can be found using the equation:

Frictional force = μk * normal force

where μk is the coefficient of kinetic friction (0.2) and the normal force is the weight of the box, which can be calculated using the equation:

Weight = mass * acceleration due to gravity

Weight = 42.2 kg * 9.81 m/s^2 = 414.78 N

Frictional force = 0.2 * 414.78 N = 82.956 N

Since the force of friction opposes the motion of the box and acts in the opposite direction of the pulling force, the magnitude of the horizontal component of the pulling force will be reduced by the frictional force.

The net force acting on the box in the horizontal direction is the difference between the horizontal component of the pulling force and the frictional force:

Net force = F_horizontal - Frictional force
Net force = 155.02 N - 82.956 N
Net force = 72.064 N

Finally, we can use Newton's second law of motion to calculate the acceleration of the box:

Net force = mass * acceleration

Acceleration = Net force / mass
Acceleration = 72.064 N / 42.2 kg
Acceleration = 1.71 m/s^2

Therefore, the acceleration of the box is 1.71 m/s^2.

To find the acceleration of the box, we need to analyze the forces acting on it. There are two main forces at play: the applied force by the student and the force of friction.

1. Resolve the applied force into its horizontal and vertical components:
- The horizontal component = 179 N * cos(29.8°)
- The vertical component = 179 N * sin(29.8°)

2. Calculate the force of friction opposing the motion:
- The force of friction = µk * (mass of the box * acceleration due to gravity)
- µk is the coefficient of kinetic friction (0.2)
- The mass of the box is given as 42.2 kg
- Acceleration due to gravity is 9.81 m/s^2

3. Since the box is moving, the force of friction will act in the opposite direction of the applied force.

4. Apply Newton's second law of motion to find the acceleration:
- Sum of forces = mass of the box * acceleration
- Forces to consider are the horizontal component of the applied force and the force of friction (acting in the opposite direction).
- Set up the equation: acceleration = (horizontal component of the applied force - force of friction) / mass of the box

Now, let's calculate the values:

Horizontal component of the applied force:
179 N * cos(29.8°) = 158.22 N

Vertical component of the applied force:
179 N * sin(29.8°) = 88.53 N

Force of friction:
0.2 * (42.2 kg * 9.81 m/s^2) = 82.52 N

Now, substituting the values into the equation for acceleration:
acceleration = (158.22 N - 82.52 N) / 42.2 kg = 0.761 m/s^2

Therefore, the acceleration of the box is 0.761 m/s^2.