Remember that f=ma and integrating a twice, you obtain the displacement
1. a block with mass m slides on a horizontal surface. the block is intially at x=0 with velocity v0.
the frictional force that the block experience is proportional to v^(3/2)
such that
F(v)=-kv^(3/2)
for positive constant K. what is the max distance that the block travel?
2. a particle of mass m is under the influences of a force F(x) that attracts the particle according to
F(x) = -kx^(-2)
where x is the distance from a fixed origin at x=0 and k is a positive constant
if the particle is released from the rest at x=d, how long does it take to reach the origin?
So and [mat]v^{-3/2}dv= \frac{k}{m}[/tex]. You can integrate that to find v(t), determining the constant of integration by the condition that . You can then set v(t)= 0 to find when it stops. Integrate again to find x(t), using to determine the constant of integration, and calculate
Now you have . That has three variables, t, x, and v. But you can use a very handy method called "quadrature" to reduce that.2. a particle of mass m is under the influences of a force F(x) that attracts the particle according to
F(x) = -kx^(-2)
where x is the distance from a fixed origin at x=0 and k is a positive constant
if the particle is released from the rest at x=d, how long does it take to reach the origin?
By the chain rule, . But so .
Now the differential equation is or . Integrate to find v as a function of x, using the fact that x(t= 0)= d, v(t= 0)= 0 which is the same as v(x=d)= 0, to determine the constant of integration.
Then solve the (separable) first order differential equation with x(0)= d.