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For a moving object, the force acting on the object varies directly with the object's acceleration. When a force of 9N acts on a certain object, the acceleration of the object is 3 m/s2 . If the force is changed to 6N , what will be the acceleration of the object?

Sagot :

MrRoyal

Answer:

The acceleration is 2m/s^2

Step-by-step explanation:

Given

Variation: Direction Variation

Represent Force with F and Acceleration with A

When

[tex]F= 9N; A=3m/s^2[/tex]

Required

Find A when F =6N

The variation can be represented as:

[tex]F \alpha A[/tex]

Convert to an equation

[tex]F = kA[/tex]

Where

[tex]k = variation\ constant[/tex]

When

[tex]F= 9N; A=3m/s^2[/tex]

We have:

[tex]9N = k * 3m/s^2[/tex]

Solve or k

[tex]k = \frac{9N}{3m/s^2}[/tex]

[tex]k = 3kg[/tex]

When

[tex]F = 6N[/tex]

We have:

[tex]F = kA[/tex]

Substitute 6N for F and 3kg for k

[tex]6N = 3kg * A[/tex]

Solve for A

[tex]A = \frac{6N}{3kg}[/tex]

[tex]A = 2m/s^2[/tex]

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