Answer:
a) -1.4 m/s^2
b) 4.1 m/s
Explanation:
Tension force that must be applied on the rope is greater than 387 N
The weighing of material is 449 N
The vertical height between the bundle and the ground = 6.1 m
a)
the mass of the bundle weighing over acceleration,
F_g = ma
m= F_g/a = 449/9.8 = 45.81 Kg
the forces that act on the bundle are the tension ( for example, upward)
and the weighing force, but they are opposite to each other so,
T-mg = ma
F_g =ma
then,
T-mg = ma
so acceleration is:
a= T-mg/m = (287-45.81×9.81)/45.81 = -1.4 m/s^2
a= -1.4 m/s^2
Minus sign indicates that the acceleration in downward direction.
b) The displacement of the bundle to reach the ground is Δy =6.1 m
v^2=u^2+2aΔy
initial velocity is zero u= 0
v^2 = 2aΔy
[tex]v= \sqrt{2\times(-1.4)(-6.1)}[/tex]
v= 4.1 m/s
Answer:
(a) 1.53 [tex]m/s^{2}[/tex]
(b) 4.06 m
Solution:
Maximum tension, [tex]T_{max} = 387 N[/tex]
Weight of the material, w = 449 N
Distance from the ground, d = 6.1 m
Now,
(a) To calculate the magnitude of the acceleration of the bundle:
The mass of the bundle:
w = mg = 449
[tex]m = \frac{449}{g} = \frac{449}{9.8} = 45.82 kg[/tex]
Now, balancing the forces:
[tex]ma = w - T_{max}[/tex]
[tex]a = \frac{w - T_{max}}{m}[/tex]
[tex]a = \frac{449 - 387}{45.82} = 1.353 m/s^{2}[/tex]
Now,
(b) The speed of the bundle can be calculated as:
Using the third eqn of motion:
[tex]v'^{2} = v^{2} + 2ad[/tex]
where
v = initial velocity = 0
v' = velocity of the bundle with which it hit the ground.
[tex]v'^{2} = 0 + 2\times 1.353\times 6.1 = 16.508[/tex]
[tex]v' = \sqrt{16.508} = 4.06\ m/s[/tex]
