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Questions and Answers
What is the magnitude of the resultant force when Mr. Williamson pulls with 1200 N at 35° and Sarah pulls with 400 N at 13°?
What is the magnitude of the resultant force when Mr. Williamson pulls with 1200 N at 35° and Sarah pulls with 400 N at 13°?
- About 1500 N
- About 1270 N
- About 1465 N (correct)
- About 1340 N
The force exerted by the student driver who crashed through 3 concrete pillars was less than 6000 N.
The force exerted by the student driver who crashed through 3 concrete pillars was less than 6000 N.
True (A)
How much force did the lineman exert on Daniel Jones?
How much force did the lineman exert on Daniel Jones?
360 N
The coefficient of friction between Mr. Williamson's tires and the road is represented as _____.
The coefficient of friction between Mr. Williamson's tires and the road is represented as _____.
Match the following scenarios with their respective forces exerted:
Match the following scenarios with their respective forces exerted:
What is the net force when the six graders pull with 12500 N and the seniors pull with 12000 N?
What is the net force when the six graders pull with 12500 N and the seniors pull with 12000 N?
If Santa weighs 2600 N, the sled will move farther with Santa in it due to increased mass.
If Santa weighs 2600 N, the sled will move farther with Santa in it due to increased mass.
What is the acceleration of the box when pulled with a force of 185 N at an angle of 25.0°?
What is the acceleration of the box when pulled with a force of 185 N at an angle of 25.0°?
Flashcards
Force
Force
The force an object exerts on another object due to its acceleration. It is calculated by multiplying the object's mass by its acceleration. For example, if a 10 kg object accelerates at 2 m/s², the force exerted is 20 Newtons.
Friction
Friction
The force that opposes the motion of an object in contact with a surface. It is affected by the coefficient of friction and the normal force. For example, when pushing a box across the floor, friction is the opposing force. If the box is heavier, the friction force will be stronger.
Normal Force
Normal Force
The force that acts perpendicular to a surface when an object is in contact with it. It is equal in magnitude but opposite in direction to the weight of the object when the surface is horizontal.
Acceleration
Acceleration
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Hooke's Law
Hooke's Law
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Resultant Force
Resultant Force
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Inertia
Inertia
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Mass
Mass
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Study Notes
Problem 1
- Given: Force 1 = 1200 N, angle 1 = 35°, Force 2 = 400 N, angle 2 = 13°
- Find: Resultant force magnitude and direction
- Method: Vector addition (resolve forces into x and y components, sum x and y components, calculate resultant magnitude and angle)
Problem 2
- Given: Mass = 1800 kg, initial velocity = 15 mph, distance = 33 meters
- Find: Force exerted by the driver
- Method: Use the work-energy theorem (Work = change in kinetic energy). Convert 15 mph to m/s. Calculate work done to stop the car. Force can be calculated as Work/distance.
Problem 3
- Given: Mass of lineman = 120 kg, acceleration = 3 m/s²
- Find: Force exerted on Danny
- Method: Use Newton's second law (F = ma)
Problem 4
- Given: Applied force = 550 N, car weight = 950 kg
- Find: Coefficient of friction
- Method: Use Newton's second law and the friction formula (F_friction = coefficient of friction * normal force). Normal force is equal to the car's weight (in this case).
Problem 5
- Given: Tension on ropes = 255 N, angles = 38°
- Find: Mass of traffic light
- Method: Resolve the tension forces into x and y components. The vertical components of tension must balance the weight of the light.
Problem 6
- Given: Force 1 = 12500 N, angle 1 = 5°, Force 2= 12000 N
- Find: Net force and winner of tug-of-war
- Method: Vector addition (resolve forces into x and y components, sum x and y components, calculate resultant magnitude and direction) The team with greater net force wins.
Problem 7
- Given: Mass = 1630 kg, horizontal force = 2970 N, time (part a) = 5.2 seconds, time (part b) = 3.9 seconds, mass of Santa = 2600 N
- Find: Distance traveled in parts a and b
- Method: Use Newton's second law to find acceleration first and then use the kinematic equations to solve for the distance.
Problem 8
- Given: Mass = 38 kg, incline = 11°, force = 118 N
- Find: Normal force, velocity in X direction, distance = 28 meters
- Method: Resolve forces into components to find normal force and acceleration. Use kinematic equations (find velocity)
Problem 9
- Given: Force = 185 N, angle = 25°, mass = 35 kg, coefficient of kinetic friction = 0.27
- Find: Acceleration of the box
- Method: Resolve force into components to find the net force acting on the box in the x direction. Calculate the friction force. Use Newton's second law (F= ma) to calculate acceleration.
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