Answer:
196 J
Explanation:
From the law of conservation of energy,
ΔU + ΔK = 0 where ΔU = change in potential energy = mgΔy and ΔK = change in kinetic energy.
ΔK = -ΔU
If we take the top of the ledge as y₁ = 5 and the ground as y₂ = 0, then Δy = y₂ - y₁ = 0 - 5 = -5 m
ΔK = -ΔU = -mgΔy = -mg -5 = 5mg = 5 m × 4 kg × 9.8 m/s = 196 J
ΔK = K₂ - K₁ where K₁ = initial kinetic of 4 kg ball = 0 J (since it is initially at rest on the ledge) and K₂ = final kinetic energy of 4 kg ball before it hits the ground.
ΔK = K₂ - K₁ = 196 J
K₂ - 0 = 196 J
K₂ = 196 J
So, it has a kinetic energy of 196 J before it hits the ground
Answer:
Explanation:
Given:
Mass, M = 4 kg
Height, S = 5 m
Initial velocity, u = 0 m/s (at rest)
Using equation of motion,
V^2 = u^2 + 2a × S
= 0 + 2 × 9.8 × 5
= 98
V = sqrt(98)
= 9.889 m/s
= 9.9 m/s
Kinetic energy, E = 1/2 × mass × v2
Note: initial kinetic energy = 0 because velocity = 0 m/s(at rest)
Final kinetic energy, Ef = 1/2 × 4 × 9.9^2
= 196 J
= 0.196 kJ
