A block with mass m=1.50 kg is initially at rest on a horizontal frictionless surface at x =0 , where x is the horizontal coordinate. A horizontally directed force is then applied to the block. The force is not constant: instead, its magnitude as a function of position is described by the relationship F(x)=(α-βx 2 )i , where x is given in units of meters, α = 2.50 N , β = 1.00 N/m2 , and i is the unit vector in the x direction.

a) What is the kinetic energy of the block as it passes through x=2.00 m?

b) What is the maximum speed of the block in the interval during which it moves from its initial position to x=2.00 m?

The force as a function of position is given above. Since the force is a function of position, we can assume that we will use work-energy theorem.

[tex]W = \Delta K = K_2 - K_1\\\int\limits^2_0 {F(x)} \, dx = \frac{1}{2}mv_2 - 0\\\int\limits^2_0 {(2.5-x^2)} \, dx = (2.5x - \frac{x^3}{3})\left \{ {{x=2} \atop {x=0}} \right. = (5 - 8/3) = 7/3[/tex]

Therefore, the kinetic energy of the block is 2.33 J.

(b) In order to find the maximum speed in this interval, we need to investigate the acceleration of the block. Since acceleration is the derivative of velocity, velocity is at its maximum when acceleration is zero.

From Newton's Second Law:

[tex]F = ma\\a(x) = F(x)/m = 1.66 - 0.66x^2[/tex]

In order this to be zero:

[tex]1.66 - 0.66x^2 = 0\\x = 1.58~m[/tex]

The velocity of the block at x = 1.58 m can be found by work-energy theorem.

[tex]\int\limits^{1.58}_0 {(2.5-x^2)} \, dx = \frac{1}{2}(1.5)v^2\\ (2.5x-\frac{x^3}{3})\left \{ {{x=1.58} \atop {x=0}} \right. = 2.63 J\\2.63 = \frac{1}{2}(1.5)v^2\\v = 1.87~m/s[/tex]