Question

At t = 0 a 4.0 kg object is passing the origin, moving right with an initial speed of 1.0 m/s. Consider the following two cases: Case A: A force F(x) = (1.0N/m) x to the right acts on the object for 4.0 m. Case B: A force F(t) = (1.0 N/s) t to the right acts on the object for 4.0 s. What is the change in kinetic energy for the object?

Answer 1

Answer:

Case A:

Work-energy theorem states that the total work done on an object is equal to the change in the kinetic energy of the object.

$W_{total} = \int{F(x)} \, dx = \int\limits^4_0 {x} \, dx = \frac{x^2}{2}\left \{ {{x = 4} \atop {x=0}} \right. = \frac{16 - 0}{2} = 8~J$

Case B:

In this case, we cannot calculate the total work done by $W = \int{F(t)} \, dx$, because the force is a function of time, not position.

So, we need to calculate the initial and final kinetic energies.

$K_1 = \frac{1}{2}mv^2 = \frac{1}{2}4(1^2) = 2~J$

For the final kinetic energy, we need to find the final velocity. What we have is the force as a function of time. From Newton’s Second Law we can find the acceleration as a function time:

$a(t) = F(t)/m = \frac{(1.0 N/s)t}{4} = \frac{t}{4}~m/s^2$

$v(t) = \int{a(t)} \, dt = \int{\frac{1}{4}t} \, dt = \frac{t^2}{8}$

From this function, at t = 4.0 s, v = 2 m/s.

$K_2 = \frac{1}{2}mv_2^2 = \frac{1}{2}4(2^2) = 8~J$

So, the change in the kinetic energy is

$\Delta K = K_2 - K_1 = 8 - 2 = 6~J$