You have landed on an unknown planet, Newtonia, and want to know what objects will weigh there. You find that when a certain tool is pushed on a frictionless horizontal surface by a 12.2 N force, it moves 16.1 m in the first 2.40 s , starting from rest. You next observe that if you release this tool from rest at 11.0 m above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia? What would it weigh on Earth?

The weight [tex]W[/tex] is the force with which a planet attracts an object with mass [tex]m[/tex] , due to the acceleration of gravity [tex]g[/tex] of said planet, and is given by the following equation:

[tex]W=m.g[/tex] (1)

Knowing this, let's find the weight of an object on Newtonia and on Earth:

1. On Newtonia:

The weight in this planet is given by (1):

[tex]W_{N}=m.g_{N}[/tex] (2)

In addition, we are told that when a certain tool is pushed on a frictionless horizontal surface by a 12.2 N force, it moves 16.1 m in the first 2.40 s , starting from rest ([tex]X_{o}=0[/tex], [tex]V_{o}=0[/tex] and [tex]t_{o}=0[/tex]).

By Newton's 2nd law of motion we know the force is given by:

[tex]F=m.a[/tex] (3) Being [tex]m[/tex] the mass of the object and [tex]a[/tex] its acceleration.

Isolating [tex]m[/tex]: [tex]m=\frac{F}{a}[/tex] (4)

We already know [tex]F=12.2N[/tex], we need to find [tex]a[/tex]. In order to approach it, we will use the following equations related to the motion in one dimension:

[tex]V=\frac{\Delta X}{\Delta t}=\frac{X-X_{o}}{t-t_{o}}[/tex] (5)

[tex]V=V_{o}+a.t[/tex] (6)

From (5):

[tex]V=\frac{16.1m-0}{2.40s-0}=6.708m/s[/tex] (7)

Substituting (7) in (6):

[tex]6.708m/s=0+a(2.40s)[/tex] (8)

Finding [tex]a[/tex]: [tex]a=\frac{6.708m/s}{2.40s}=2.795m/s^{2}[/tex] (9)

Substituting (9) in (4):

[tex]m=\frac{12.2n}{2.795m/s^{2}}[/tex] (10)

[tex]m=4.364kg[/tex] (11)

On the other hand, we are told that if you release this tool from rest ( [tex]V_{oy}=0[/tex] and [tex]t_{o}=0[/tex]) at [tex]y=11.0m[/tex] above the ground, it takes [tex]t=2.58s[/tex] to reach the ground.

In this case we will use the following equation:

[tex]y=V_{oy}.t-\frac{1}{2}g_{N}.t^{2}[/tex] (12)

[tex]11m=0-\frac{1}{2}g_{N}(2.58s)^{2}[/tex] (13)

Finding [tex]g_{N}[/tex]:

[tex]g_{N}=-6.610m/s^{2}[/tex] (14) Where the negative sign only indicates the acceleration is "downwards", to the center of the planet.

Substituting (14) in (1):

[tex]W_{N}=(4.364kg)(6.610m/s^{2})[/tex] (15)

[tex]W_{N}=28.846N[/tex] This is the weight of the tool on Newtonia.

2. On Earth:

In the case of our planet, we will use the same weight equation, knowing the acceleration due gravity on Earth is [tex]g_{E}=9.8m/s^{2}[/tex]:

[tex]W_{E}=m.g_{E}[/tex] (16)

[tex]W_{E}=(4.364kg)(9.8mm/s^{2})[/tex] (17)

Finally:

[tex]W_{E}=42.767N[/tex] This is the weight of the same tool on Earth.