Answer:
Vp = 1 [m/s]
Explanation:
In order to solve this problem, we must use the principle of conservation of the amount of movement. In the same way, analyze the before and after of the actions.
<u>The moment before</u>
The 50kg person is still hold (no movement) with the 2kg helmet
<u>The moment after</u>
The helmet moves at 25[m/s] in one direction, the person moves in the opposite direction, due to the launch of the helmet.
In this way we can apply the principle of conservation of movement, expressing the before and after. To the left we have the before and to the right of the equal sign we have the after.
Σm*V1 = Σm*V
where:
m = total mass = (2 + 50) = 52[kg]
V1 = velocity before the lunch = 0
(50 + 2)*V1 = (25*2) - (Vp*50)
0 = 50 - 50*Vp
50 = 50*Vp
Vp = 1 [m/s]
Answer:
d = <23, 33, 0> m
, F_W = <0, -9.8, 0>
, W = -323.4 J
Explanation:
We can solve this exercise using projectile launch ratios, for the x-axis the displacement is
x = vox t
Y Axis
y =
t - ½ g t²
It's displacement is
d = x i ^ + y j ^ + z k ^
Substituting
d = (23 i ^ + 33 j ^ + 0) m
Using your notation
d = <23, 33, 0> m
The force of gravity is the weight of the body
W = m g
W = 1 9.8 = 9.8 N
In vector notation, in general the upward direction is positive
W = (0 i ^ - 9.8 j ^ + 0K ^) N
W = <0, -9.8, 0>
Work is defined
W = F. dy
W = F dy cos θ
In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º
Cos 180 = -1
W = -F y
W = - 9.8 (33-0)
W = -323.4 J
F(g)= Gm1m2/ r^2
If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)