Answer
(a) See the free body diagram attached
(b) $F_f = 62.2~N$
(c) $F_N = 199~N$
(d) $F_p = 99.8~N$
Work Step by Step
(a) The free body diagram is attached.
(b) Since the mower is moving at constant speed, the horizontal component of the person's force $F_p$ must be equal in magnitude to the friction force.
$F_f = F_p ~cos(\theta) = (88.0~N)~cos(45^{\circ})$
$F_f = 62.2~N$
(c) The normal force is equal in magnitude to the sum of the weight plus the vertical component of the person's applied force $F_p$.
$F_N = mg + F_p~sin(\theta)$
$F_N = (14.0~kg)(9.80~m/s^2) + (88.0~N)~sin(45^{\circ})$
$F_N = 199~N$
(d) We can calculate the required acceleration.
$a = \frac{v}{t} = \frac{1.5~m/s}{2.5~s} = 0.60~m/s^2$
We can use a force equation to find the required force which the person needs to apply.
$\sum F = ma$
$F_p ~cos(\theta) - F_f = ma$
$F_p = \frac{ma+F_f}{cos(\theta)} = \frac{(14.0~kg)(0.60~m/s^2 + 62.2~N)}{cos(45^{\circ})}$
$F_p = 99.8~N$
