Question

A 5 kg object is moving in a straight-line with an initial speed of v m/s. It takes 13 s for the speed of the object to increase to 13 m/s and it kinetic energy increases at a rate of 15 J/s. What is the initial speed v (in m/s)?

Answer 1

The object's kinetic energy changes according to

dK/dt = 15 J/s

If v is the object's initial speed, then its initial kinetic energy is

K (0) = 1/2 (5 kg) v ²

Use the fundamental theorem of calculus to solve for K as a function of time t :

$K(t) = K(0) + \displaystyle\int_0^t \left(15\frac{\rm J}{\rm s}\right)\,\mathrm du = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)t$

After t = 13 s, the object's kinetic energy is

K (13 s) = 1/2 (5 kg) (13 m/s)² = 422.5 J

Put this as the left side in the equation above for K(t) and solve for v :

$422.5\,\mathrm J = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)(13\,\mathrm s)$

==>   v ≈ 9.5 m/s