Question

Using a rope that will snap if the tension in it exceeds 387 N, you need to lower a bundle of old roofing material weighing 449 N from a point 6.1 m above the ground. Obviously if you hang the bundle on the rope, it will snap. So, you allow the bundle to accelerate downward. (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

Answer 1

Answer:

(a) $a\approx1.4 m.s^{-2}$

(b) $v\approx 4.133 m.s^{-1}$

Explanation:

Given:

• Limiting tension of snapping of the rope, T= 387 N

• Weight of the object to be lifted, $w=449 N$

• ∴mass, $\Rightarrow m= 45.8163 kg$

• height of letting down the weight, h = 6.1 m

(a)

Now,

The force to be compensated for  being on the verge of snapping:

(T-w) = 62 N

Therefore, we need to produce and acceleration equivalent to the above force.

∵$F=m.a$

$62=45.8163\times a$

$a= \frac{62}{45.8163}$

$a\approx 1.4 m.s^{-2}$

(b)

From the equation of motion ,we have:

$v^{2} =u^{2} +2a.s$....................(2)

where:

u= initial velocity= 0 (here, starting from rest)

v= final velocity = ?

$a= 1.4 m.s^{-2}$

s= displacement =h =6.1 m

Now, putting the values in eq. (2)

$v^2= 0^2 + 2\times 1.4\times 6.1$

$v\approx 4.133 m.s^{-1}$        is the velocity with which the body will hit the ground in the given conditions.