University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

200

Pages:

5
Author:

Jabin F.

#Some Basic Concepts of Mechanics:- Impulse = change in momentum Force, F = dp/dt avg. force, F_avg = ∫F dt / t J = area under force-time graph with t-axis a = dv/dt v = ∫a dt work done, dW = F · dl P = dW/dt = F · v avg. power = total work done / total time Q. A body of mass m falls from rest at a height 'h' under gravity through a dense medium which provides a resistive force F = -kv², where k is a constant and v is the speed. It will hit the ground with a kinetic energy: Solution: F_net = mg - kv² = ma use a = dv/dx mg - kv² = m * dv/dx * dx ∫(mg - kv²) dx = ∫m * dv -∫(mg - kv²) dx = ∫m * v dv [mgx - (k/3)v³] = (1/2)mv² Solve for v² and substitute into KE = (1/2)mv² to find the final kinetic energy. Continuation of the previous problem: ∫(-kx/h) dx = ∫m ln(mg - kv²) dv -∫kx dx = m[ln(mg - kv²) - ln(mg)] -kh = m ln[(mg - kv²) / mg] e^(-kh/m) = (mg - kv²) / mg kv² = mg(1 - e^(-kh/m)) v² = (mg/k)(1 - e^(-2kh/m)) Then kinetic energy, KE = (1/2)mv² = (mg² / 2k)(1 - e^(-2kh/m)) 2. A particle travels in a medium along a horizontal linear path. The initial velocity of the particle is u₀, and the viscous force acting on it is proportional to its instantaneous velocity. In the absence of any other force, which one of the following figures correctly represents the velocity of the particle as a function of time? (a) (b) (c) (d) F = -kv = ma a = dv/dt -kv = m * dv/dt -k/m * ∫dt = ∫(1/v) dv -kt/m = ln(v/v₀) v = v₀ * e^(-kt/m)