Question

1. a 45 n force is applied to a 5.0 kg object as shown. if the coefficient of friction is 0.55, what is the acceleration of the object? 6. a 450 n force is applied to a 64 kg object as shown. if the coefficient of friction is 0.35, what is the acceleration of the object? 7. an object begins sliding down a ramp. if the object was initially at rest 1.5 m from the base of the ramp and the coefficient of friction is 0.30, how long does it take for the object to reach the bottom of the ramp? bonus 8. the coefficient of static friction between a 25 kg object and a surface is 0.55. determine the minimum force needed to move the object from rest if the force is applied at an angle of 35° above the horizontal. answers: 2. 4.6m/s² clown 4. 0.28s 6. 1.90m/s² right 8. you tell me 1. 190n 24°safe 3. 1.51m/s² down 5. 4.68m/s² right 7. 1.52s

Explanation:

Step1: Analyze vertical forces for question 5

For the 5.0 - kg object, the weight $F_g=mg = 5\times9.8 = 49N$. The vertical component of the applied force $F_{A\perp}=45\sin40^{\circ}$. The normal force $N = F_g - F_{A\perp}=49 - 45\sin40^{\circ}$.

Step2: Calculate frictional force

The frictional force $f=\mu N=\mu(49 - 45\sin40^{\circ})$, where $\mu = 0.55$.

Step3: Analyze horizontal forces and use Newton's second - law

The horizontal component of the applied force $F_{A\parallel}=45\cos40^{\circ}$. According to Newton's second - law $F_{net}=ma$, so $F_{A\parallel}-f = ma$. Substitute $f$ and solve for $a$:
\[

$$\begin{align*} a&=\frac{45\cos40^{\circ}-\mu(49 - 45\sin40^{\circ})}{5}\\ &=\frac{45\times0.766-0.55\times(49 - 45\times0.643)}{5}\\ &=\frac{34.47-0.55\times(49 - 28.935)}{5}\\ &=\frac{34.47-0.55\times20.065}{5}\\ &=\frac{34.47 - 11.03575}{5}\\ &=\frac{23.43425}{5}\\ & = 4.68685\approx4.68m/s^{2} \end{align*}$$

\]

Step4: Analyze vertical forces for question 6

For the 64 - kg object, the weight $F_g=mg=64\times9.8 = 627.2N$. The vertical component of the applied force $F_{A\perp}=450\sin25^{\circ}$. The normal force $N = F_g+F_{A\perp}=627.2 + 450\sin25^{\circ}$.

Step5: Calculate frictional force

The frictional force $f=\mu N=\mu(627.2 + 450\sin25^{\circ})$, where $\mu = 0.35$.

Step6: Analyze horizontal forces and use Newton's second - law

The horizontal component of the applied force $F_{A\parallel}=450\cos25^{\circ}$. According to Newton's second - law $F_{net}=ma$, so $F_{A\parallel}-f = ma$. Substitute $f$ and solve for $a$:
\[

$$\begin{align*} a&=\frac{450\cos25^{\circ}-\mu(627.2 + 450\sin25^{\circ})}{64}\\ &=\frac{450\times0.9063-0.35\times(627.2 + 450\times0.4226)}{64}\\ &=\frac{407.835-0.35\times(627.2+189.17)}{64}\\ &=\frac{407.835-0.35\times816.37}{64}\\ &=\frac{407.835 - 285.7295}{64}\\ &=\frac{122.1055}{64}\\ &\approx1.90m/s^{2} \end{align*}$$

\]

Step7: Analyze forces for question 7

The net force along the ramp is $F_{net}=mg\sin\theta-\mu mg\cos\theta$. According to Newton's second - law $F_{net}=ma$, so $a = g\sin\theta-\mu g\cos\theta$. Given $\theta = 24^{\circ}$, $\mu = 0.30$, $g = 9.8m/s^{2}$, we have $a=9.8\times\sin24^{\circ}-0.30\times9.8\times\cos24^{\circ}=9.8\times(0.4067-0.30\times0.9135)=9.8\times(0.4067 - 0.27405)=9.8\times0.13265\approx1.3m/s^{2}$. Using the kinematic equation $x = v_0t+\frac{1}{2}at^{2}$, since $v_0 = 0$ and $x = 1.5m$, we have $1.5=\frac{1}{2}\times1.3t^{2}$, then $t^{2}=\frac{1.5\times2}{1.3}=\frac{3}{1.3}\approx2.3077$, and $t=\sqrt{2.3077}\approx1.52s$.

Step8: Analyze forces for bonus question 8

The normal force $N = mg - F\sin35^{\circ}$, where $m = 25kg$, $g = 9.8m/s^{2}$. The maximum static frictional force $f_s=\mu_sN=\mu_s(mg - F\sin35^{\circ})$. The horizontal component of the applied force is $F\cos35^{\circ}$. At the verge of moving, $F\cos35^{\circ}=f_s=\mu_s(mg - F\sin35^{\circ})$. Rearranging the equation:
\[

$$\begin{align*} F\cos35^{\circ}&=\mu_smg-\mu_sF\sin35^{\circ}\\ F\cos35^{\circ}+\mu_sF\sin35^{\circ}&=\mu_smg\\ F&=\frac{\mu_smg}{\cos35^{\circ}+\mu_s\sin35^{\circ}} \end{align*}$$

\]
Substitute $\mu_s = 0.55$, $m = 25kg$, $g = 9.8m/s^{2}$:
\[

$$\begin{align*} F&=\frac{0.55\times25\times9.8}{0.8192+0.55\times0.5736}\\ &=\frac{0.55\times25\times9.8}{0.8192 + 0.31558}\\ &=\frac{134.75}{1.13478}\\ &\approx119N \end{align*}$$

\]

Answer:

1. $4.68m/s^{2}$ [right]
2. $1.90m/s^{2}$ [right]
3. $1.52s$
4. $119N$