Question

Two packages at UPS start sliding down the$20°$ramp shown in FIGURE $P7.33$. Package A has a mass of$5.0kg$ and a coefficient of friction of $0.20$. Package B has a mass of $10kg$ and a coefficient of friction of$0.15.$ How long does it take package A to reach the bottom?

Short answer

The time required by the package to reach the bottom is$0.7sec$.

Step-by-step solution

Given information

Mass of block A, $m_A=5kg$

Mass of block B, $m_B=10kg$

Coefficient of friction of block A, ${\mu }_A=0.2$

Coefficient of friction of block B, ${\mu }_B=0.15$

The angle of inclination, $\varphi =20°$

Distance of the inclination,$s=2m$

Explanation

The force which opposes the motion of block A, is the friction force

$$\begin{gathered} F_{friction}={\mu }_A{m}_{A}g\cos 20° \\ {F}_{friction}=\left(0.2\right)\left(5kg\right)\left(9.81m/s^2\right)\left(\cos \left(20°\right)\right) \\ =9.21N \end{gathered}$$

The force which tends to move to block A in the forward direction is calculated as

$$\begin{gathered} F_{applied}=m_{A}g\sin 20°+m_{B_{}}g\sin 20° \\ =\left(5kg\right)\left(9.81m/s^2\right)\left(\sin \left(20°\right)\right)+\left(10kg\right)\left(9.81m/s^2\right)\left(\sin \left(20°\right)\right) \\ =50.32N \end{gathered}$$

The net force on the block,

$$\begin{gathered} F_{net}=\left(50.32N\right)-\left(9.21N\right) \\ =41.11N \end{gathered}$$

Acceleration of block A is calculated as

$$\begin{gathered} F_{net}=m_{A}a \\ 41.11N=\left(5kg\right)\times a \\ a=8.22m/s^2 \end{gathered}$$

The time required by block A to reach the bottom is calculated as

$s=ut+\frac{1}{2}a{t}^2$

$$\begin{gathered} 2m=0+\frac{1}{2}\times \left(8.22m/s^2\right)\times t^2 \\ t=0.69s \\ t=0.7s \end{gathered}$$

The time required by package A to reach the bottom is $0.7sec$.