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Kisachek [45]
2 years ago
5

David and Fiona want to know about how mass affects the motion of an object. David uses three objects with masses of 2 kg, 3 kg,

and 8 kg, writes out his procedure, and records his observations on the force needed to move the objects. Fiona is investigating three different objects using observation. What is the difference in the way they approached this question?
A. Only Fiona gathered evidence from her investigation.
B. David used scientific articles for evidence.
C. David's observations can be replicated by repeating his procedure.
D. Fiona's observations can be replicated by observing different objects.
Physics
1 answer:
Natali [406]2 years ago
8 0

Answer: the answer is D if its not i'm really sorry

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Answer:

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Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following parallel forces acting on the block (in parallel direction to the direction of movement):

F = applied force.

Fr = friction force.

wx = parallel component of the weight.

According to Newton´s second law:

∑F = m · a

Where "m" is the mass of the block and "a" its acceleration.

Then:

F - Fr - wx = m · a

Since the block is to be pushed at a constant velocity, the acceleration is zero. Then:

F - Fr - wx = 0

F = Fr + wx

The applied force has to be equal to the friction force plus the parallel component of the weight to push the block at constant velocity.

The friction force is calculated as follows:

Fr = μ · N

Where N is the normal force and μ is the coefficient of friction.

Notice that the normal force is of the same magnitude as the perpendicular component of the weight, wy.

Let´s apply Newton´s second law in the perpendicular direction to show this:

∑F = m · a

N - wy = m · a

The acceleration of the block in the perpendicular direction is zero. Then:

N - wy = 0

N = wy

And wy can be obtained by trigonometry (see figure):

wy = W · cos θ

N = wy = mg ·  cos θ

The parallel component of the weight is calculated using trigonometry (see figure):

wx = W · sin θ

wx = mg · sin θ

Then the applied force will be:

F = Fr + wx

F = μ · N + mg · sin θ      (N = wy = mg ·  cos θ)

F = μ · mg ·  cos θ  + mg · sin θ

F = mg ( μ · cos θ + sin θ)

The correct answer is the b.

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