ll.Problem Solving:
Directions Answer problems and show complete solution and box the final answer.
Take note : The quantitative relationship between work and two forms of mechanical energy is expressed by the following equation: KE,+PE,+W=KE,+PE,
Where:KE,=Initial kinectic Energy
KE,=Final Kinetic Energy
PE,=Initial Potential Energy
PE,=Final Potential Energy
W=Work Done
Problem #1:
Consider the falling and rolling motion of the ball in the following two resistance-free situations. In one situation ,the ball falls off the top of the platform to the floor.In the other situations,the ball rolls from the top of the platform along the staircase-like pathway to the floor.
a.For each situation,indicate what types of forces are doing work upon the ball.
a.1.The ball falls off the top of the platform to the floor:____________________________
b.indicate whether the energy of the ball is conserved and explain why_______________________________________
c.finally,fill in the blanks for the 2-kg.Show your solution
c1.KE=_A_ J
C2.V=_B_J
C3.PE_C_J
C4.KE_D_J
C5.V=_E_J
c6.PE=_F_J
c7.KE=_G_J
C8.v=_H_m/s
c9.PE=_I_J
C10.KE_J_J
C11.v=_k_m/s
patulong naman po ibrainliest ko makasagot
Directions Answer problems and show complete solution and box the final answer.
Take note : The quantitative relationship between work and two forms of mechanical energy is expressed by the following equation: KE,+PE,+W=KE,+PE,
Where:KE,=Initial kinectic Energy
KE,=Final Kinetic Energy
PE,=Initial Potential Energy
PE,=Final Potential Energy
W=Work Done
Problem #1:
Consider the falling and rolling motion of the ball in the following two resistance-free situations. In one situation ,the ball falls off the top of the platform to the floor.In the other situations,the ball rolls from the top of the platform along the staircase-like pathway to the floor.
a.For each situation,indicate what types of forces are doing work upon the ball.
a.1.The ball falls off the top of the platform to the floor:____________________________
b.indicate whether the energy of the ball is conserved and explain why_______________________________________
c.finally,fill in the blanks for the 2-kg.Show your solution
c1.KE=_A_ J
C2.V=_B_J
C3.PE_C_J
C4.KE_D_J
C5.V=_E_J
c6.PE=_F_J
c7.KE=_G_J
C8.v=_H_m/s
c9.PE=_I_J
C10.KE_J_J
C11.v=_k_m/s
patulong naman po ibrainliest ko makasagot
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Answer:
A. KE₁ = 50 J
B. v₁ = 7.0711 m/s
C. PE₂ = 50J
D. KE₂ = 50 J
E. v₂ = 7.0711 m/s
F. PE₄ = 0
G. KE₄ = 100J
H. v₄ = 10 m/s
I. PE₃ = 0
J. KE₃ = 100 J
K. v₃ = 10 m/s
Explanation:
Let us solve using The principle of energy conservation which states that energy is neither created nor destroyed.
(KE₀ + PE₀) = (KE₁ + PE₁)
since: KE₀ = 0 J
PE₀ = 100J
PE₁ = 50J (since it is half-way down (or height is half))
(0 + 100) = (KE₁ + 50)
KE₁ = 100 - 50
A. KE₁ = 50 J
since: KE = (1/2)mv²
50 J = (1/2)(2kg)v² [since joules(J) is kg(m²/s²)]
2( 50 kg m²/s² ) / (2kg) = v₁²
v₁² = 50 m²/s²
v₁ = √(50 m²/s²)
B. v₁ = 7.0711 m/s
the 3rd instance is when the ball has same height from the 2nd instance/figure.
so:
C. PE₂ = PE₁ = 50J
D. KE₂ = KE₁ = 50 J
E. v₂ = v₁ = 7.0711 m/s
solving the 4th instance when the ball is at the floor or height = 0
(KE₀ + PE₀) = (KE₃ + PE₃)
since: height is 0
PE= mgh = mg(0) = 0
I. PE₃ = 0
(0 + 100) = (KE₃ + 0)
J. KE₃ = 100 J
KE = (1/2)mv²
100 J = (1/2)(2kg)v₃²
[(2)(100 kg m²/s²)] / (2kg) = v₃²
v₃² = 100 m²/s²
v₃ = √(100 m²/s²)
K. v₃ = 10 m/s
let's solve for the dropped ball:
(KE₀ + PE₀) = (KE₄ + PE₄)
again: height is 0
PE= mgh = mg(0) = 0
F. PE₄ = 0
(0 + 100) = (KE₄ + 0)
G. KE₄ = 100J
KE = (1/2)mv²
100 J = (1/2)(2kg)v₄²
[(2)(100 kg m²/s²)] / (2kg) = v₄²
v₄² = 100 m²/s²
v₄ = √(100 m²/s²)
H. v₄ = 10 m/s
== kindly check my answer if it's correct. And I hope this will help you. Have fun and study hard. =) ==