Respuesta :
Answer:
[tex]w'=5.679\ N[/tex] on the planet
[tex]w=22.43\ N[/tex] on earth
Explanation:
Given:
- initial velocity of the tool before pushing, [tex]u=0\ m.s^{-1}[/tex]
- force applied on the tool, [tex]F=12\ N[/tex]
- displacement of the tool, [tex]s=16.4\ m[/tex]
- time taken for the displacement, [tex]t=2.5\ s[/tex]
- height of releasing the tool, [tex]h=10.3\ m[/tex]
- time taken by the tool to fall on the ground, [tex]t_v=2.88\ s[/tex]
Now using the equation of motion:
[tex]s=u.t+\frac{1}{2}a.t^2[/tex]
where:
a = acceleration of the object
[tex]16.4=0+0.5\times a\times 2.5^2[/tex]
[tex]a=5.248\ m.s^{-2}[/tex]
Now the mass of the tool:
[tex]m=\frac{F}{a}[/tex]
[tex]m=\frac{12}{5.248}[/tex]
[tex]m=2.2866\ kg[/tex]
Using the equation of motion when the tool is dropped:
[tex]h=u.t_v+\frac{1}{2} \times g.t_v^2[/tex]
here:
g = acceleration due to gravity on the planet
[tex]10.3=0+0.5\times a\times 2.88^2[/tex]
[tex]g=2.4836\ m.s^{-2}[/tex]
Weight of the tool in the planet:
[tex]w'=m.g[/tex]
[tex]w'=2.2866\times 2.4836[/tex]
[tex]w'=5.679\ N[/tex]
Weight of the tool on the earth:
[tex]w=m.g'[/tex]
[tex]w=2.2866\times 9.81[/tex]
[tex]w=22.43\ N[/tex]
Answer:
Explanation:
Force, F = 12 N
distance, s = 16.4 m
time, t = 2.5 s
initial velocity, u = 0 m/s
Let a be the acceleration
use second equation of motion
s = ut + 1/2 at²
16.4 = 0 + 0.5 x a x 2.5 x 2.5
a = 5.25 m/s²
Let m be the mass of the tool.
F = ma
12 = m x 5.25
m = 2.286 kg
Now for vertical motion
h = 10.3 m
u = 0 m/s
t = 2.88 m/s²
Let the acceleration due to gravity on that planet is g'
Use second equation of motion
10.3 = 0 + 0.5 x g' x 2.88 x 2.88
g' = 2.485 m/s²
Let W' be the weight of the tool on that planet
W' = m g' = 2.286 x 2.485 = 5.68 N
Let W be the weight of the tool on the earth
W = m x g = 2.286 x 9.8 = 22.4 N
