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A 5 kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant. The mass of the planet is unknown. After 2 s, the object has fallen 30 m. Air resistance is considered to be negligible. What is the gravitational force exerted on the 5 kg object near the planet's surface?

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Final answer:

The gravitational force exerted on the 5 kg object near the planet's surface is 75 N.

Step-by-step explanation:

The gravitational force exerted on the 5 kg object near the planet's surface can be calculated using Newton's law of universal gravitation. The formula to calculate the gravitational force is:

F = (G * m1 * m2) / r^2

Where F is the gravitational force, G is the gravitational constant (6.67 × 10^-11 N m^2/kg^2), m1 is the mass of the planet, m2 is the mass of the object, and r is the distance between the center of the planet and the object.

In this case, we know that the object falls 30 m in 2 s, so we can calculate the acceleration using the formula:

a = (2 * d) / t^2

Where a is the acceleration, d is the distance fallen (30 m), and t is the time taken (2 s).

Using the acceleration, we can calculate the force using Newton's second law:

F = m * a

Substituting the given values, we get:

F = 5 kg * (30 m / 2 s^2)

Solving this equation, we find that the gravitational force exerted on the 5 kg object near the planet's surface is 75 N.

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