Answer
(a) $7.2J$
(b) $11m/s$
(c) $17.8J$
Work Step by Step
(a) We can find the required kinetic energy as
$K.E=\frac{1}{2}mv^2$
We plug in the known values to obtain:
$K.E=\frac{1}{2}(0.40)(6.0)^2$
$K.E=7.2J$
(b) We can find the required speed at $t=2.0s$ as follows:
$v=\sqrt{\frac{2K.E}{m}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{2(25)}{0.40}}$
$v=11m/s$
(c) The required work done can be determined as
$W=\Delta K.E=K.E_f-K.E_i$
We plug in the known values to obtain:
$W=25J-7.2J$
$W=17.8J$
