Answer
(a) v = 4.96 m/s
(b) The final velocity is 4.96 m/s, which is the same result that we calculated with the work-energy theorem.
Work Step by Step
(a) $W = K_2- K_1$
$F~d = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$
$mv_1^2 + 2F~d = mv_2^2$
$v_2^2 = \frac{mv_1^2 + 2F~d}{m}$
$v_2 = \sqrt{\frac{mv_1^2 + 2F~d}{m}}$
$v_2 = \sqrt{\frac{(7.00~kg)(4.00~m/s)^2 + (2)(10.0~N)(3.0~m)}{7.00~kg}}$
$v_2 = 4.96~m/s$
(b) $F = ma$
$a = \frac{F}{m} = \frac{10.0~N}{7.00~kg} = 1.43~m/s^2$
We can use kinematics to find the final speed $v$.
$v^2 = v_0^2 + 2ax$
$v = \sqrt{(4.00~m/s)^2 + (2)(1.43~m/s^2)(3.0~m)}$
$v = 4.96~m/s$
As expected, the final velocity is 4.96 m/s, which is the same result that we calculated with the work-energy theorem.
