Question

1. the system below is accelerating at 2.9 m/s² as shown. determine the mass, m. (dont ask mrs. kiel. you can do this question. follow the steps. it is just like the other ones only now you are solving for a mass not acceleration or tension!) answers a = 4.29m/s² right t₁ = 138n t₂ = 138n 2. a = 4.46m/s² d/le t₁=t₂ = 26.8n 3. a = 4.72m/s² d/l t₁ = t₂ = 102n 4. m = 111

Explanation:

Step1: Analyze forces on 6.0 - kg block

For the 6.0 - kg block on the inclined plane, the normal force $F_N$ is given by $F_N = mg\cos\theta$, where $m = 6.0$ kg, $g=9.8$ m/s² and $\theta = 20^{\circ}$. So $F_N=6\times9.8\times\cos20^{\circ}\approx 55.4$ N. The frictional force $F_f=\mu F_N$, with $\mu = 0.70$, so $F_f = 0.70\times55.4\approx38.8$ N. The component of the gravitational force along the incline is $F_{g\parallel}=mg\sin\theta=6\times9.8\times\sin20^{\circ}\approx 20.2$ N. Let the tension in the string be $T$. According to Newton's second - law $T - F_f - F_{g\parallel}=ma$, where $a = 2.9$ m/s². So $T=ma+F_f + F_{g\parallel}=6\times2.9+38.8 + 20.2=17.4+38.8+20.2 = 76.4$ N.

Step2: Analyze forces on mass $m$

For the mass $m$, according to Newton's second - law $mg - T=ma$. We can re - arrange the formula to solve for $m$: $m=\frac{T}{g - a}$. Substitute $T = 76.4$ N, $g = 9.8$ m/s² and $a = 2.9$ m/s² into the formula. $m=\frac{76.4}{9.8 - 2.9}=\frac{76.4}{6.9}\approx11.1$ kg.

Answer:

$m\approx11.1$ kg