Answer
(a) The horizontal force is 613 N.
(b) (i) W = 0
(ii) W = 549 J
Work Step by Step
(a) We can find the angle $\theta$ the rope makes with the vertical.
$sin(\theta) = \frac{2.0~m}{3.5~m}$
$\theta = arcsin(\frac{2.0~m}{3.5~m})$
$\theta = 34.8^{\circ}$
vertical forces:
$T~cos(34.8^{\circ}) = mg$
$T = \frac{mg}{cos(34.8^{\circ})}$
horizontal forces:
$F = T~sin(34.8^{\circ})$
$F = mg~tan(34.8^{\circ})$
$F = (90.0~kg)(9.80~m/s^2)~tan(34.8^{\circ})$
$F = 613~N$
The horizontal force is 613 N.
(b) (i) Since the rope exerts a force at a $90^{\circ}$ angle to the direction of motion, the work done by the rope is zero.
(ii) We can use the angle to find the change in height $h$ of the bag.
$\frac{2.0~m}{3.5~m-h} = tan(\theta)$
$2.0~m = (3.5~m-h)~tan(\theta)$
$h = \frac{(3.5~m)~tan(34.8^{\circ})-2.0~m}{tan(34.8^{\circ})}$
$h = 0.622~m$
We can use the change in height to find the work done by the worker.
$Work = mgh$
$Work = (90.0~kg)(9.80~m/s^2)(0.622~m)$
$Work = 549~J$
