If the motor exerts a force of f = (600 + 2s2) n on the cable, determine the speed of the 100-kg crate when it rises to s = 15 m. the crate is initially at rest on the ground.

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

When you go to ask a qustion there will be a picture of a paper clip looking thing and then it will pop up your files and you click on the imajes and then you up load it

The law of conservation of momentum will be the same no matter where you apply it.the equation will be  77.4 + m (mass of canister) *3.6