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A 120 kg crate is on a ramp that is inclined at 15°. What is the gravitational force component acting parallel to the ramp?

a) 294.3 N
b) 490.5 N
c) 112.8 N
d) 98.1 N

Answer :

To calculate the gravitational force component parallel to the ramp for a 120 kg crate on a 15 degree incline, use the formula F_parallel = mg sin(15 degrees). The result is approximately 306.48 N, which doesn't exactly match any of the provided options, suggesting a rounding or calculation discrepancy.None of the options are correct .

The question asks for the gravitational force component acting parallel to the ramp on which a 120 kg crate rests inclined at 15°.

To find this component, we will use the formula for the parallel force component: Fparallel = mg sin(θ), where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s²), and θ is the angle of the incline.

Here's the calculation:

Fparallel = 120 kg * 9.8 m/s² * sin(15°)

Fparallel ≈ 120 kg * 9.8 m/s² * 0.2588

Fparallel ≈ 306.48 N

The closest answer choice to this calculated value is (b) 294.3 N, but it should be noted that the provided options do not include the exact calculated value, which could mean there may be a slight discrepancy due to rounding of the angle's sine value or a difference in gravitational acceleration used.

None of the options are correct .

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