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Kinetic Energy and Work in AP Physics 1

Created by @nathanedwards
 at November 1st 2023, 10:02:27 am.
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#work
#kinetic-energy
#ap-physics-1

AP Physics 1 Exam Question:

A 2 kg block is initially at rest on a frictionless table. A constant force of 10 N is then applied to the block at an angle of 30° above the horizontal. The force is applied over a distance of 5 m. Calculate the work done on the block by the applied force, and determine its change in kinetic energy.

Solution:

To calculate the work done on the block by the applied force, we can use the equation:

Work=Force⋅Distance⋅cos⁡(θ) \text{Work} = \text{Force} \cdot \text{Distance} \cdot \cos(\theta)

where the force is the component of the applied force in the direction of motion, the distance is the distance over which the force is applied, and theta (θ) is the angle between the force and the direction of motion.

In this case, the force applied is 10 N. To find the component of the force in the direction of motion, we can multiply the force applied by the cosine of the angle:

Forceparallel=Force⋅cos⁡(θ) \text{Force}_\text{parallel} = \text{Force} \cdot \cos(\theta)
Forceparallel=10 N⋅cos⁡(30°) \text{Force}_\text{parallel} = 10 \, \text{N} \cdot \cos(30°)
Forceparallel=10 N⋅32 \text{Force}_\text{parallel} = 10 \, \text{N} \cdot \frac{\sqrt{3}}{2}
Forceparallel=53 N \text{Force}_\text{parallel} = 5 \sqrt{3} \, \text{N}

The distance over which the force is applied is given as 5 m.

Plugging the values into the equation for work done:

Work=(53 N)⋅(5 m)⋅cos⁡(30°) \text{Work} = (5 \sqrt{3} \, \text{N}) \cdot (5 \, \text{m}) \cdot \cos(30°)
Work=(53 N)⋅(5 m)⋅32 \text{Work} = (5 \sqrt{3} \, \text{N}) \cdot (5 \, \text{m}) \cdot \frac{\sqrt{3}}{2}
Work=752 J \text{Work} = \frac{75}{2} \, \text{J}

Therefore, the work done on the block by the applied force is 37.5 J.

To determine the change in kinetic energy, we can use the work-energy theorem, which states:

Work=ΔKE \text{Work} = \Delta \text{KE}

Thus, the change in kinetic energy is equal to the work done on the block:

ΔKE=37.5 J \Delta \text{KE} = 37.5 \, \text{J}

Therefore, the change in kinetic energy of the block is 37.5 J.

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