Using Newton\'s second law (F = ma) and integarting between the initial and fina
ID: 1500065 • Letter: U
Question
Using Newton's second law (F = ma) and integarting between the initial and final positions, show the work-energy relation holds: where T_1, T_2 are the initial and final kinetic energies, and W = f? F-dx is the work done by the force Two small bodies of mass m are connected as drawn, where B is constrained by a guide to only move vertically. The configuration is initially at rest, and the only force is that of gravity. Using work and energy, find the velocity of A as it crosses the vertical line.Explanation / Answer
1. Let's apply a constant force F to a mass m as it moves, in one dimension, a distance x. (It might, for instance, be the magnetic force that we used in our section on Newton's laws.) The force is constant over x, x increases linearly over x, so the work done F.dx increases linearly over x
The total work done on the mass is
where F is the total force acting on the mass. Subsituting for Newton's second law, F = ma = m(dv/dt), gives:
where we have written the mutliplication and division explicitly. dx, dv and dt are all small quantities, but there is no reason why we cannot change the order of multiplication. So let's write:
The advantage of this rearrangment is that we can now do the integral easily:
Suppose we start from v = 0, then the total work done to accelerate mass m from rest to a speed v is:
This quantity is so useful that we give it a name, the kinetic energy and write
2. Energy = 1/2 * m*v^2 = mgh . In this equation m is the mass. v is the velocity you want to calculate, g is gravitational acceleration (= 9.8m/s/s) and h is the distance you say you know. You see that m gets canceled out , hence
v = (2*g*h)
This is in the case of a projectile hitting the ground from a height of h.