Question

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?

Short answer

1. The magnitude of the bundle’s acceleration will put the rope on the verge of snapping is $1.35\mathrm{m}/{\mathrm{s}}^2$.
2. The hitting speed of the bundle at that acceleration is 4.1 m/s.

Step-by-step solution

Given information

• 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, $∆y=6.1\mathrm{m}$.

Understanding the concept of force on the object

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:

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

Draw the free body diagram

(a) Calculate the magnitude of the bundle’s acceleration

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.

(b) Calculate the hitting speed of the bundle

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.