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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 30m. Air resistance is considered to be negligible. What is the gravitational force exerted on the 5kg5⁢kg object near the planet’s surface?

(A) 5 N
(B) 15 N
(C) 37.5 N
(D) 75 N

Respuesta :

mauricioalessandrell

Answer:

The gravitational force exerted on the object is 75 N (answer D)

Explanation:

Hi there!

The gravitational force is calculated as follows:

F = m · g

Where:

F = force of gravity.

m = mass of the object.

g = acceleration due to gravity (unknown).

For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · a · t²

Where:

y = position at time t.

y0 = initial position.

v0 = initial velocity.

t = time.

g = acceleration due to gravity.

Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.

Then, after 2 seconds, the height of the object will be -30 m:

y = y0 + v0 · t + 1/2 · g · t²

-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²

-30 m = 1/2 · g · 4 s²

-30 m = 2 s ² · g

-30 m/2 s² = g

g = -15 m/s²

Then, the magnitude of the gravitational force will be:

F = m · g

F = 5 kg · 15 m/s²

F = 75 N

The gravitational force exerted on the object is 75 N (answer D)

Have a nice day!

nuhulawal20

The gravitational force exerted on the object near the planet’s surface is 75 Newton.

Hence, Option (D) 75N is the correct answer.

Given the data in the question;

  • Mass of object; [tex]m = 5kg[/tex]

Since the object was released from rest

  • Initial velocity; [tex]u = 0[/tex]
  • Distance of the fall; [tex]s = 30m[/tex]
  • Time taken; [tex]t = 2s[/tex]

First we determine the acceleration due to gravity of the planet using the Second Equation of Motion:

[tex]s = ut + \frac{1}{2} at^2\\\\[/tex]

Where s is the distance, u is the initial velocity, a is acceleration and t is time.

We substitute our values into the equation

[tex]30m=[ 0*2s ] + [ \frac{1}{2} * a * (2s)^2]\\\\30m= \frac{1}{2} * a * 4s^2\\\\30m = a* 2s^2\\\\a = \frac{30m}{2s^2}\\\\a = 15m/s^2[/tex]

Now, to calculate the gravitational force exerted on the object, we use Newton's Second Law of Motion:

[tex]F = ma[/tex]

Where m is the mass of the object and a is acceleration due to gravity.

We substitute our values into the equation

[tex]F = 5kg\ *\ 15m/s^2\\\\F = 75 kg,m/s^2\\\\F = 75N[/tex]

The gravitational force exerted on the object near the planet’s surface is 75 Newton.

Hence, Option (D) 75N is the correct answer.

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