Using a rope that will snap if the tension in it exceeds387 N you need to lower a bundle of old roofing material weighing449 Nfrom a point 6.1 mabove the ground. (a)What magnitude of the bundle’s acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground?

• The weight of the bundle is,$W=449\mathrm{N}$.
• The tension in the rope is, T = 387N.
• The vertical distance of the bundle from the ground is, $∆y=6.1\mathrm{m}$.

Newton’s second law of motion states that the force is equal to the product of mass and the acceleration of the object. Tension is the form of force. The free-body diagram shows all the forces acting on the body and their directions.

Draw a free-body diagram. By using Newton’s second law we can find the magnitude of the acceleration of the bundle. By using the kinematic equation, we can find the hitting speed of the bundle on the ground.

Formulae:

${\overrightarrow{F}}_{net}=m\overrightarrow{a}$

The mass of the bundle is

$$\begin{gathered} W=mg \\ m=\frac{W}{g} \\ =\frac{449\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =45.8\mathrm{kg} \end{gathered}$$

By applying Newton’s second law along y-axis as,

$$\begin{gathered} {\overrightarrow{F}}_{net}=m\overrightarrow{a} \\ T-mg=-ma \\ a=-\frac{T-mg}{m} \\ =-\frac{387\mathrm{N}-449\mathrm{N}}{45.8\mathrm{kg}} \\ =1.35\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the magnitude of acceleration is 1.35 m.

The bundle is hitting at the ground from the maximum height hence its initial speed is. We can use the sign convention according to the direction of motion of the bundle. The acceleration and displacement of the bundle are negative.

By applying the third kinematical equation along the vertical axis as

$$\begin{gathered} v^2=v_0^2+2a∆y \\ =v_0^2+2a∆y \\ =2a∆y \\ v=\sqrt{2a∆y} \\ =\sqrt{2\times {(-1.35\mathrm{m}/\mathrm{s})}^2\times (-6.1\mathrm{m})} \\ =4.1\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the acceleration is 4.1 m/s.