Question

1. a boy pushes a lawn mower ($m = 17.9$ kg) starting from rest across a horizontal lawn by applying a force of 32.9 n straight along the handle, which is inclined at an angle of 35.1° above the horizontal. the magnitude of the mowers acceleration is 1.37 m/s², which lasts for 0.58 s, after which the mower moves at a constant velocity. determine the magnitude of (a) the normal force on the mower (b) the frictional force on the mower (c) the maximum velocity of the mower (d) the force applied by the boy needed to maintain the constant velocity

Explanation:

Step1: Analyze vertical forces for normal force

The vertical - component of the applied force is $F_{y}=F\sin\theta$, and the weight of the mower is $mg$. According to Newton's second law in the vertical direction ($\sum F_y = 0$ during the acceleration phase and also when moving at constant velocity), $N + F\sin\theta=mg$. So, $N = mg - F\sin\theta$.
Given $m = 17.9$ kg, $g = 9.8$ m/s², $F = 32.9$ N, and $\theta = 35.1^{\circ}$.
$N=17.9\times9.8-32.9\times\sin(35.1^{\circ})$
$N = 175.42-32.9\times0.576$
$N = 175.42 - 18.95$
$N\approx156.5$ N

Step2: Analyze horizontal forces for frictional force

According to Newton's second law in the horizontal direction during the acceleration phase ($\sum F_x=ma$), $F\cos\theta - f=ma$. So, $f = F\cos\theta - ma$.
Given $F = 32.9$ N, $\theta = 35.1^{\circ}$, $m = 17.9$ kg, and $a = 1.37$ m/s².
$F\cos\theta=32.9\times\cos(35.1^{\circ})=32.9\times0.818 = 26.91$ N
$ma=17.9\times1.37 = 24.52$ N
$f=26.91 - 24.52=2.39$ N

Step3: Calculate maximum velocity

Using the kinematic equation $v = v_0+at$. Since $v_0 = 0$ m/s, $a = 1.37$ m/s², and $t = 0.58$ s.
$v=0 + 1.37\times0.58=0.795$ m/s

Step4: Determine force for constant - velocity motion

When the mower moves at a constant velocity, the net force in the horizontal direction is zero. So, $F_{new}\cos\theta=f$.
We know $f = 2.39$ N and $\cos\theta=\cos(35.1^{\circ}) = 0.818$.
$F_{new}=\frac{f}{\cos\theta}=\frac{2.39}{0.818}\approx2.92$ N

Answer:

(a) $156.5$ N
(b) $2.39$ N
(c) $0.795$ m/s
(d) $2.92$ N