Answer
(a) $3.5$
(b) $24.528m/s^2$
(c) $0.24528m/s$
Work Step by Step
(a) We know that
$W=mg$
$\implies W=(67Kg)(9.81m/s^2)$
$W=656.6N$
The required ratio can be determined as
$ratio=\frac{F}{W}$
We plug in the known values to obtain:
$ratio=\frac{2300}{656.6}=3.5$
(b) We can find the magnitude and direction of the required acceleration as follows:
$\Sigma F_y=ma_y$
$\implies 2300N+(67Kg)(-9.8m/s^2)=(67Kg)a_y$
$\implies a_y=\frac{2300N-656.6N}{67Kg}$
$a_y=24.528m/s^2$
(c) We know that
$\Delta v_y=a_y t$
We plug in the known values to obtain:
$\Delta v_y=(24.528m/s^2)(0.01s)$
$\Delta v_y=0.24528m/s$
