Question

question 5.3a: the figure shows a box of mass m = 4.75 kg pulled to the right across a horizontal surface by a constant - tension force of magnitude t = 20.5 n. the tension force is inclined at an angle θ = 24.0° above the horizontal and the friction force has magnitude f = 12.5 n. the box is pulled a distance d = 1.80 m.

1. find a
2. find μk

Explanation:

Step1: Analyze forces in y - direction

In the vertical direction, the net force is zero since there is no acceleration in the y - direction. So, $N + T\sin\theta=mg$, where $N$ is the normal force, $T$ is the tension force, $\theta$ is the angle of the tension force with the horizontal, $m$ is the mass of the box, and $g = 9.8\ m/s^{2}$. We can solve for $N$: $N=mg - T\sin\theta$.
$m = 4.75\ kg$, $T = 29.5\ N$, $\theta = 24^{\circ}$, $g=9.8\ m/s^{2}$
$N=4.75\times9.8-29.5\times\sin(24^{\circ})$
$N = 46.55-29.5\times0.4067$
$N = 46.55 - 12.007$
$N\approx34.54\ N$

Step2: Analyze forces in x - direction

In the horizontal direction, according to Newton's second - law $F_{net,x}=ma_x$. The net force in the x - direction is $T\cos\theta - f=ma$, where $f=\mu_kN$ is the frictional force.
$T\cos\theta-\mu_kN = ma$
We know that $T = 29.5\ N$, $\theta = 24^{\circ}$, $N\approx34.54\ N$, $f = 12.5\ N$
First, find $T\cos\theta$: $T\cos\theta=29.5\times\cos(24^{\circ})=29.5\times0.9135 = 26.95\ N$

Sub - step 2.1: Find acceleration $a$

From $T\cos\theta - f=ma$, we can solve for $a$:
$a=\frac{T\cos\theta - f}{m}$
$a=\frac{26.95 - 12.5}{4.75}$
$a=\frac{14.45}{4.75}$
$a = 3.04\ m/s^{2}$

Sub - step 2.2: Find $\mu_k$

Since $f=\mu_kN$, then $\mu_k=\frac{f}{N}$
$\mu_k=\frac{12.5}{34.54}\approx0.362$

Answer:

1. $a = 3.04\ m/s^{2}$
2. $\mu_k\approx0.362$