Answer:
mb = 3.75 kg
Explanation:
System of forces in balance
ΣFx =0
ΣFy = 0
Forces acting on the box
T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left
T₂ = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.
Wb :Weightt of the box (vertical downward)
x-y T₁ and T₂ components
T₁x= T₁cos50°
T₁y= T₁sin50°
T₂x= 30*cos75° = 7.76 N
T₂y= 30*sin75° = 28.98 N
Calculation of the Wb
ΣFx = 0
T₂x-T₁x = 0
T₂x=T₁x
7.76 = T₁cos50°
T₁ = 7.76 /cos50° = 12.07 N
ΣFy = 0
T₂y+T₁y-Wb = 0
28.98 + 12.07(cos50°) = Wb
Wb = 36.74 N
Calculation of the mb ( mass of the box)
Wb = mb* g
g: acceleration due to gravity = 9.8 m/s²
mb = Wb/g
mb = 36.74 /9.8
mb = 3.75 kg
Given:
m = 4 kg, the mass of the object
h = 5 m, distance fallen
Neglect air resistance.
The PE (potential energy) is
PE = mgh = (4 kg)*(9.8 m/s²)*(5 m) = 196 J
The PE is converted into KE (kinetic energy) after the fall.
Therefore the PE decreased by 196 J ≈ 200 J
Answer: d. It has decreased by 200 J
work done by gravitational force = mass × g × height
= mgh
= 5 × 10 × 8 N
<h3>= 400 N</h3>
