Respuesta :
Explanation:
The given data is as follows.
[tex]v_{o}[/tex] = 0, [tex]x_{o}[/tex] = 0, [tex]t_{o}[/tex] = 0
[tex]x_{1}[/tex] = 17.0 m, [tex]t_{1}[/tex] = 2.10 sec
As the force P is constant and the mass "m" of the tool is constant then it means that the acceleration "a" will also be constant.
Now,
[tex]x_{1} = x_{o} v_{o}t_{1} + \frac{1}{2}at^{2}_{1}[/tex]
= [tex]0 + (0)(2.10 s) + \frac{1}{2}a(2.10)^{2}[/tex]
17.0 = 2.205a
a = [tex]7.71 m/s^{2}[/tex]
Also, we know that
F = [tex]\frac{m}{a}[/tex]
m = [tex]\frac{F}{a}[/tex]
So, m = [tex]\frac{12.8 N}{7.71 m/s^{2}}[/tex]
= 1.66 kg
Since, the tool is subject to its weight W and is in free fall. Hence,
[tex]x_{1} = x_{o} + v_{o}t_{1} + \frac{1}{2}gt^{2}_{1}[/tex]
10.0 m = [tex]0 + (0)(2.88 s) + \frac{1}{2} \times g \times (2.88s)^{2}[/tex]
g = 2.411 [tex]m/s^{2}[/tex]
Hence, weight of tool in Newtonia is as follows.
W = mg
= [tex]1.66 kg \times 2.411 m/s^{2}[/tex]
= 4.00 N
Hence, weight of the tool on Newtonia is 4.00 N.
And, weight of the tool on the Earth is as follows.
W = [tex]2.411 \times 9.80[/tex]
= 23.62 N
Hence, weight of the tool on Earth is 23.62 N.
