Question

A 5kg 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 2s, the object has fallen 30m. Air resistance is considered to be negligible. What is the gravitational force exerted on the 5kg object near the planet’s surface?

Answer 1

Answer:

75 N

Explanation:

t = Time taken = 2 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 30 m

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0\times 2+\frac{1}{2}\times a\times 2^2\\\Rightarrow a=\frac{30\times 2}{2^2}\\\Rightarrow a=15\ m/s^2$

The acceleration due to gravity on the planet is 15 m/s²

Force

F = ma

$F=5\times 15\\\Rightarrow F=75\ N$

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

Answer 2

We have that for the Question it can be said that he gravitational force exerted on the 5kg object near the planet’s surface

F=75N

From the question we are told

A 5kg 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 2s, the object has fallen 30m.

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

Generally the equation for the Motion   is mathematically given as

$s=ut+1/2at^2\\\\Therefore\\\\30=0+1/2(a)(2)^2\\\\a=15m/s^2$

Therefore

F=ma

F=5*15

F=75N

Hence, the gravitational force exerted on the 5kg object near the planet’s surface

F=75N

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