8) A 6.0-kg block initially at rest is pulled to the right along a frictionless, horizontal surface by a constant horizontal force of 12 N. Use the work-kinetic energy theorem to find the block’s speed after it has moved 3.0 m.
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A 6.0-kg block initially at rest is pulled to the right along a frictionless, horizontal surface by a constant horizontal force of 12 N. Use the work-kinetic energy theorem to find the block’s speed after it has moved 3.0 m.
To find the block's speed after it has moved 3.0 m using the work-kinetic energy theorem, we need to calculate the net work done on the block.
The work-kinetic energy theorem states that the work done on an object is equal to the change in its kinetic energy. Mathematically, it can be expressed as:
Work = ΔKE
Where:
Work is the net work done on the object
ΔKE is the change in kinetic energy of the object
In this case, since the block starts from rest, its initial kinetic energy (KE) is zero.
As the block is being pulled to the right by a constant horizontal force, the work done on the block is given by:
Work = Force × Distance × cos(θ)
Where:
Force is the applied force on the block
Distance is the distance the block is moved
cos(θ) is the angle between the applied force and the displacement of the block
In this case, the force is given as 12 N and the distance is given as 3.0 m. Since the force is acting horizontally, the angle between the applied force and the displacement is 0 degrees, so cos(0) = 1.
Plugging in the values, we get:
Work = (12 N) × (3.0 m) × (1)
Work = 36 Joules
Since the work done on the block is equal to the change in its kinetic energy, we can write:
36 J = ΔKE
Since the initial kinetic energy of the block is zero, the change in kinetic energy is equal to the final kinetic energy:
36 J = KEf
Now, using the equation for kinetic energy:
KE = (1/2)mv^2
Where:
KE is the kinetic energy
m is the mass of the object
v is the velocity of the object
In this case, the mass of the block is given as 6.0 kg.
Plugging in the values, we have:
36 J = (1/2)(6.0 kg)v^2
Simplifying the equation:
36 J = 3.0 kg × v^2
Dividing both sides by 3.0 kg:
12 J/kg = v^2
Taking the square root of both sides to solve for v:
√(12 J/kg) = v
v ≈ 3.46 m/s
Therefore, the block's speed after it has moved 3.0 m is approximately 3.46 m/s.
To solve this problem, we can use the work-kinetic energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy.
The work done on an object is given by the equation:
Work = Force * Distance * cos(theta)
In this case, the force applied is 12 N, and the distance moved is 3.0 m. Since the force is acting horizontally, the angle between the force and the direction of motion is 0 degrees, so cos(theta) = 1.
Therefore, the work done on the block is:
Work = 12 N * 3.0 m * 1 = 36 J
According to the work-kinetic energy theorem, this work done on the block is equal to the change in its kinetic energy. So, we can set up the equation:
Work = Change in Kinetic Energy
Since the block starts from rest, its initial kinetic energy is zero. Let's denote the final speed of the block as v. The change in kinetic energy is then given by:
Change in Kinetic Energy = (1/2) * mass * (final velocity)^2
Plugging in the given values, we have:
36 J = (1/2) * 6.0 kg * (v)^2
Simplifying the equation, we get:
36 J = 3.0 kg * (v)^2
Dividing both sides by 3.0 kg, we have:
(v)^2 = 12 J / 3.0 kg
(v)^2 = 4 m^2/s^2
Taking the square root of both sides, we get:
v = √(4 m^2/s^2)
v = 2 m/s
Therefore, the block's speed after it has moved 3.0 m is 2 m/s.