Respuesta :
The motor that exerts force is F = 600+2s²
Mass of crate is m= 100kg
Distance raises is s= 15m
Crate at rest is V₁ =0
We will solve this using principle of energy and work because involves displacement, velocity, and force
Work energy is
T₁ +ΣU₁₋₂ = T₂
Whereby T₁ = initial Kinetic energy
= 1/2mw₁² = 1/2m(o)₁² = 0 Because initial is at rest.
Motor exerts the force on the cable forces acting in the cable at C are two times motor exerts force cable M work done by the exerted motor is
U₁₋₂ = 2∫s²Fds
U₁₋₂ = 2∫s²(600+2S²)ds
=2(600×s+ 2s³/3)¹⁵
= 2(600 ×15 + 2× 15³/3)
=22500N
Work done by crate is the crates' weight acts downward to take the negative sign
U₁₋₂ = -(m × g × s)
U₁₋₂ = -(100× 9.81 × 15)
= -14715N
T₂ = Final kinetic energy = 1/2mw²₂
T₁ +∑U₁₋₂ =T₂
0 + (22500 +(-14715)) = 1/2(100) V²₂
V₂ = 12.47m/s
Mass of crate is m= 100kg
Distance raises is s= 15m
Crate at rest is V₁ =0
We will solve this using principle of energy and work because involves displacement, velocity, and force
Work energy is
T₁ +ΣU₁₋₂ = T₂
Whereby T₁ = initial Kinetic energy
= 1/2mw₁² = 1/2m(o)₁² = 0 Because initial is at rest.
Motor exerts the force on the cable forces acting in the cable at C are two times motor exerts force cable M work done by the exerted motor is
U₁₋₂ = 2∫s²Fds
U₁₋₂ = 2∫s²(600+2S²)ds
=2(600×s+ 2s³/3)¹⁵
= 2(600 ×15 + 2× 15³/3)
=22500N
Work done by crate is the crates' weight acts downward to take the negative sign
U₁₋₂ = -(m × g × s)
U₁₋₂ = -(100× 9.81 × 15)
= -14715N
T₂ = Final kinetic energy = 1/2mw²₂
T₁ +∑U₁₋₂ =T₂
0 + (22500 +(-14715)) = 1/2(100) V²₂
V₂ = 12.47m/s
Answer:
Speed of the 100 kg crate, v = 136 m/s
Explanation:
It is given that,
Force exerted by a motor, F = (600+2 s²) N
mass of the crate, m = 100 kg
We have to find the speed of the crate. Force is given by the product of mass and acceleration.
F = m a
g is acceleration due to gravity
[tex]F=m\dfrac{d^2s}{dt^2}[/tex]
[tex]\dfrac{d^2s}{dt^2}=\dfrac{600+2s^2}{100}[/tex]
[tex]\dfrac{d^2s}{dt^2}=6+0.02s^2[/tex]
[tex]\dfrac{d^2s}{dt^2}-6-0.02s^2=0[/tex]
or
s"- 6 - 0.02 s² = 0
On solving the above differential equation using calculator with when initially the crate is at rest s(0) = 0 and s'(0) = 15 m
[tex]s(t)=150\ e^{-\sqrt{2}t}+150\ e^{\sqrt{2}t}-300[/tex]
Differentiating above equation w.r.t t to get velocity of the crate.
So, v = 136 m/s
Hence, this is the required solution.
