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First Law Sample Problem
What net force is required to maintain a 5,000 kg object moving at a constant velocity of magnitude 7,500 m/s?
Solution
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Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass.
The formula for Newton's Second Law of Motion is F = ma, where F is the force applied, m is the mass of the object, and a is the acceleration produced.
This law is also known as the law of acceleration.
The direction of the force applied determines the direction of the acceleration produced.
The greater the force applied, the greater the acceleration produced, and the greater the mass of the object, the smaller the acceleration produced.
This law is used to calculate the force required to move an object of a certain mass at a certain acceleration.
A force of 1 kg·m/s 2 is renamed 1 newton
Newton's Second Law of Motion is essential in understanding the behavior of objects in motion and is used in various fields such as engineering, physics, and sports.
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Second Law Sample Problem
How much force is required to cause an object of mass 2 kg to have an acceleration of 4 m/s 2?
Solution
Fnet = ma = (2 kg)(4 m/s 2 ) = 8 N
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Friction force is a force that opposes motion between two surfaces that are in contact.
There are three types of friction:
The factors that affect friction are:
The strengths of these two types of friction forces are given by the following equations:
The weight of an object is the gravitational force exerted on it by a gravitational field.
Mass is an intrinsic property of an object that measures its inertia. An object’s mass does not change with location.
Since weight is a force, we can use F = ma to compute it, where the acceleration is the gravitational force imposed on an object. Therefore, setting a = g, the equation F = ma becomes
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F = (mv^2)/r
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The following diagrams show examples of a ball on a string traveling in a horizontal circle and a vertical circle.
Example 1
An object of mass 5 kg moves at a constant speed of 6 m/s in a circular path of radius 2 m. Find the object’s acceleration and the net force responsible for its motion.
Solution
By definition, an object moving at constant speed in a circular path is undergoing uniform circular motion. Therefore, it experiences a centripetal acceleration of magnitude v 2/r, always directed toward the center of the circle:
The force that produces the centripetal acceleration is given by Newton’s Second Law, coupled with the equation for centripetal acceleration:
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Example 2
A 10 kg mass is attached to a string that has a breaking strength of 200 N. If the mass is whirled in a horizontal circle of radius 80 cm, what maximum speed can it have? Assume the string is horizontal.
Solution
The first thing to do in problems like this is to identify what force(s) provide the centripetal force. In this example, the tension in the string (FT ) provides the centripetal force (Fc):
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Pulleys are devices that change the direction of the tension force in the cords that slide over them.
Pulley Sample Problem
In the diagram above, assume that the tabletop is frictionless. Determine the acceleration of the blocks once they’re released from rest.
Solution
There are two blocks, so draw two free-body diagrams: The positive directions for each block must coincide. If the block on the table travels to the right then the hanging block travels down. This is why down is positive for the hanging block.
To get the acceleration of each one, we use Newton’s Second Law, Fnet = ma. Notice that while Fw = FN for the block on the table, Fw does not equal FT for the hanging block, because that block is in motion in the downward direction.
Note that there are two unknowns, FT and a, but we can eliminate FT by adding the two equations, and then we can solve for a.
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An inclined plane is basically a ramp. If an object of mass m is on the ramp, then the force of gravity on the object, Fw = mg, has two components: One that’s parallel to the ramp (mg sin θ) and one that’s normal to the ramp (mg cos θ), where θ is the incline angle.
The force driving the block down the inclined plane is the component of the block’s weight that’s parallel to the ramp: mg sin θ.
When analyzing objects moving up or down inclined planes it is almost always easiest to rotate the coordinate axes such that the x-axis is parallel to the incline and the y-axis is perpendicular to the incline, as shown in the diagram.
The object would accelerate in both the xand y-directions as it moved down along the incline if you did not rotate the axis. However, with the rotated axes, the acceleration in the y-direction is zero. Now we only have to worry about the acceleration in the x-direction.
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Inclined Planes Sample Problem
A block slides down a frictionless inclined plane that makes a 30° angle with the horizontal. Find the acceleration of this block.
Solution
Let m denote the mass of the block, so the force that pulls the block down the incline is mg sin θ, and the block’s acceleration down the plane is
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