a college student takes her book to class by placing them in a box and pulling it around school. She pulls on the box with a force of 90N at a angle of 30 relative to the horizontal. the box of books has a mass of 20 kg and the coefficiant of kinetic friction between the bottom of the box and the sidwalk is 0.5. What is the net froce acting horizontally and what is the acceleration of the box.
To find the net force acting horizontally and the acceleration of the box, we need to calculate the gravitational force, the force of friction, and the net force.
First, let's calculate the gravitational force:
Gravitational force = mass * gravitational acceleration
Gravitational force = 20 kg * 9.8 m/s^2 (gravitational acceleration)
Gravitational force = 196 N
Next, let's calculate the force of friction:
Force of friction = coefficient of kinetic friction * normal force
Normal force = mass * gravitational acceleration
Normal force = 20 kg * 9.8 m/s^2
Normal force = 196 N
Force of friction = 0.5 (coefficient of kinetic friction) * 196 N
Force of friction = 98 N
Now, let's calculate the horizontal component of the pulling force:
Horizontal component of the pulling force = pulling force * cos(angle)
Horizontal component of the pulling force = 90 N * cos(30°)
Horizontal component of the pulling force = 90 N * 0.866
Horizontal component of the pulling force = 77.94 N
Finally, let's calculate the net force:
Net force = horizontal component of the pulling force - force of friction
Net force = 77.94 N - 98 N
Net force = -20.06 N (negative sign indicates opposite direction)
The net force acting horizontally is approximately -20.06 N.
Now, we can use Newton's second law (F = ma) to calculate the acceleration of the box:
Net force = mass * acceleration
-20.06 N = 20 kg * acceleration
acceleration = -20.06 N / 20 kg
acceleration = -1.003 m/s^2 (negative sign indicates opposite direction)
The acceleration of the box is approximately -1.003 m/s^2.
To find the net force acting horizontally on the box, we need to consider two forces: the force applied by the student (90N) and the force of friction.
1. The force applied by the student is given as 90N at an angle of 30 degrees relative to the horizontal. To find the horizontal component of this force, we can use trigonometry.
Horizontal Force Component = Applied Force * cos(angle)
= 90N * cos(30°)
= 90N * 0.866
= 77.94N
2. Next, we need to calculate the force of friction. The formula to calculate friction is:
Force of Friction = Coefficient of Friction * Normal Force
The normal force is equal to the weight of the box (mass x gravity).
Normal Force = mass x gravity
= 20kg x 9.8m/s²
= 196N
Force of Friction = Coefficient of Kinetic Friction * Normal Force
= 0.5 * 196N
= 98N
3. Now, we can calculate the net force acting horizontally:
Net Force = Horizontal Force Component - Force of Friction
= 77.94N - 98N
= -20.06N (Note: The negative sign indicates that the forces are in opposite directions.)
The net force acting horizontally on the box is -20.06N.
4. Finally, to find the acceleration of the box, we use Newton's second law: F = m * a, where F is the net force and m is the mass.
Net Force = mass * acceleration
Rearranging the equation, we find:
Acceleration = Net Force / mass
= -20.06N / 20kg
= -1.003 m/s² (Note: The negative sign indicates acceleration in the opposite direction of the applied force.)
The acceleration of the box is -1.003 m/s².