The answer is false. The speed of the astronaut cancels out the force of gravity, causing a 'stationary freefall'. While under these effects, it is not required for an astronaut to 'strengthen' his body.
Work = Force x Distance = 500 x 4 = 2000 Nm = 2000 J
Answer:
15009
Explanation:
PE = mgh
PE = 61.2(9.81)(10 * 2.50)
PE = 15009.3
Answer:
d = 0.05 [m] = 50 [mm]
Explanation:
We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.
![E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm]](https://tex.z-dn.net/?f=E_%7Bk%7D%3DF%2Ad%5C%5C400%20%3D%208000%2Ad%5C%5Cd%20%3D%200.05%20%5Bm%5D%20%3D%2050%20%5Bmm%5D)
Answer:
a) Em₀ = 42.96 104 J
, b)
= -2.49 105 J
, c) vf = 3.75 m / s
Explanation:
The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has
Em = K + U
a) Let's look for the initial mechanical energy
Em₀ = K + U
Em₀ = ½ m v2 + mg and
Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142
Em₀ = 36 104 + 6.96 104
Em₀ = 42.96 104 J
b) The work of the friction force is equal to the change in the mechanical energy of the body
= Em₂ -Em₀
Em₂ = K + U
Em₂ = ½ m v₂² + m g y₂
Em₂ = ½ 50 85 2 + 50 9.8 427
Em₂ = 180.625 + 2.09 105
Em₂ = 1,806 105 J
= Em₂ -Em₀
= 1,806 105 - 4,296 105
= -2.49 105 J
The negative sign indicates that the work that force and displacement have opposite directions
c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job
We have that the work of friction is equal to the change of mechanical energy
= ΔEm
= Emf - Emo
-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴
½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵
½ 50.0 vf² = 0.561
vf = √ 0.561 25
vf = 3.75 m / s