If the Kinetic energyk=wv^2/2g, find approximately the change in the Kinetic energy as w changes from 49 to 49.5 and v changes from 1600 to 1590. (Hint: Takeg= 32ft/s2)
Answers
The change in kinetic energy is approximately -4500 units.
If the kinetic energy, K = wv²/2g , where w changes from 49 to 49.5 and v changes from 1600 to 1590.
We have to find the approximately change in the kinetic energy.
We use partial derivatives’ concept here to find the change in kinetic energy.
Here, K = wv²/2g
using partial differentiation,
δK = 1/2g [2wvδv + v²δw ]
∵ w = 49, δw = 49.5 - 49 = 0.5 , v = 1600 and δv = 1590 - 1600 = -10
∴ δK = 1/2g [2 × 49 × 1600 × -10 + (1600)² × 0.5] Given, g = 32 ft/s²
⇒ δK = 1/(64) [ -1568000 + 1280000]
= -4500
Therefore the change in kinetic energy is approximately -4500 units. (in f.p.s)
Answer:
∂K = 1/2g [2wv∂ + v + v²∂W].
When,
W = 49, ∂W = 49.5 - 49 = 0.5, v = 1,600 and ∂v = 1,590 - 1,600 = -10.
Therefore,
∂K = 1/2g [2 × 49 × 1,600 × (-10) + (1,600)² × 0.5].
∂K = 1/64 [ -1568000 + 1280000 ].
∂K = -4,500.
Hence, the change in kinetic energy = -4,500.