Physics: Newton's Laws: Free Body Diagrams and Net Force
Drawing forces and calculating the motion they cause
Physics: Newton's Laws: Free Body Diagrams and Net Force
Drawing forces and calculating the motion they cause
Physics - Grade 9-12
- 1
A book rests motionless on a level table. List the forces acting on the book and describe the direction of each force.
Only include forces acting on the book, not forces the book exerts on other objects.
The forces on the book are gravity downward and the normal force upward from the table. Since the book is motionless, the two forces are equal in size and the net force is zero. - 2
A 10 kg box is pushed to the right with a force of 50 N. Friction pushes to the left with a force of 20 N. Find the net force on the box and its acceleration.
The net force is 50 N - 20 N = 30 N to the right. The acceleration is a = Fnet / m = 30 N / 10 kg = 3.0 m/s^2 to the right. - 3
A 65 kg student stands in an elevator. The scale pushes up on the student with a normal force of 750 N. The student's weight is 637 N. Find the net force and acceleration of the student.
Treat upward as positive and downward as negative.
The net force is 750 N upward - 637 N downward = 113 N upward. The acceleration is a = 113 N / 65 kg = 1.7 m/s^2 upward. - 4
A sled is pulled horizontally with a force of 80 N to the right. The sled moves at constant velocity across the snow. What is the net force on the sled, and what must the friction force be?
Constant velocity means acceleration is zero.
The net force is zero because the sled moves at constant velocity. The friction force must be 80 N to the left to balance the pulling force. - 5
A skydiver falls straight downward at terminal velocity. Describe the free-body diagram and the net force on the skydiver.
The free-body diagram has weight downward and air resistance upward. At terminal velocity, these forces are equal in size, so the net force is zero. - 6
A 2.0 kg cart has three horizontal forces acting on it: 12 N right, 7 N left, and 3 N left. Find the net force and acceleration.
The total leftward force is 7 N + 3 N = 10 N. The net force is 12 N - 10 N = 2 N to the right, so the acceleration is 2 N / 2.0 kg = 1.0 m/s^2 to the right. - 7
A 12 kg box is on a frictionless ramp inclined at 30 degrees. Find the component of the box's weight parallel to the ramp and describe the direction of the acceleration.
On an incline, the component parallel to the ramp is mg sin(theta).
The parallel component of weight is mg sin(30 degrees) = 12 kg x 9.8 m/s^2 x 0.5 = 58.8 N down the ramp. Since there is no friction, the box accelerates down the ramp. - 8
A 20 kg crate is pushed across a floor with a 70 N force to the right. Kinetic friction is 35 N to the left. Find the net force and acceleration of the crate.
The net force is 70 N - 35 N = 35 N to the right. The acceleration is a = 35 N / 20 kg = 1.75 m/s^2 to the right. - 9
Two students pull on a cart. One pulls 18 N east and the other pulls 12 N west. If friction is negligible, what is the net force on the cart?
Opposite directions subtract.
The net force is 18 N east - 12 N west = 6 N east. The cart will accelerate east if it has mass. - 10
A lamp hangs at rest from a ceiling by a single cord. The lamp has a mass of 4.0 kg. Find the tension in the cord.
The lamp is at rest, so the net force is zero. The tension equals the lamp's weight: T = mg = 4.0 kg x 9.8 m/s^2 = 39.2 N upward. - 11
A 1,200 kg car slows down with an acceleration of 4.0 m/s^2 opposite its motion. Find the net braking force on the car.
A force opposite the motion can produce a negative acceleration.
The net force is Fnet = ma = 1,200 kg x 4.0 m/s^2 = 4,800 N opposite the car's motion. This force is backward if the car is moving forward. - 12
A free-body diagram for a soccer ball in flight after being kicked shows only one force: gravity downward. Air resistance is ignored. Is this diagram correct, and why?
Forces require an interaction that is happening at that moment.
Yes, the diagram is correct if air resistance is ignored. After the ball leaves the foot, the kick is no longer acting on it, so gravity is the only force on the ball. - 13
A 5 kg object is pulled upward by a rope with a tension of 60 N. The object's weight is 49 N. Find the net force and acceleration.
The net force is 60 N upward - 49 N downward = 11 N upward. The acceleration is a = 11 N / 5 kg = 2.2 m/s^2 upward. - 14
In an Atwood machine, a 5.0 kg mass and a 3.0 kg mass hang over a frictionless pulley. Find the magnitude of the acceleration of the system.
For the whole system, the net driving force is the difference between the two weights.
The unbalanced force is the difference in the weights: (5.0 kg - 3.0 kg) x 9.8 m/s^2 = 19.6 N. The total mass is 8.0 kg, so the acceleration is 19.6 N / 8.0 kg = 2.45 m/s^2. - 15
A 6.0 kg block is pulled to the right across a table by a 40 N force. The block accelerates at 3.0 m/s^2 to the right. Find the friction force acting on the block.
The applied force and friction combine to make the net force.
The net force must be Fnet = ma = 6.0 kg x 3.0 m/s^2 = 18 N to the right. Since the applied force is 40 N to the right, friction must be 40 N - 18 N = 22 N to the left.