A stone of mass 50 gram is released from a rubber catapult that has been streched through 4cm. Is the force constant of the rubber is 500N/m calculate the velocity which the stone leaves that catapult

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oeerivona oeerivona · Answer 1 · 2021-01-17T04:38:42+01:00

Answer:

The velocity with which the stone leaves the catapult is 4 m/s

Explanation:

The mass of the stone released, m = 50 g = 0.05 kg

The length by which the rubber catapult is stretched, Δx = 4 cm = 0.04 m

The force constant of the rubber, k = 500 N/m

By the conservation of energy formula, we have;

K₁ + U₁ = K₂ + U₂

Where;

K₁ = The initial kinetic energy of the stone and the catapult system = 0 J (The stone and the rubber of the catapult are held back stationary)

U₁ = The initial potential energy of the stone and the catapult system = 1/2·k·(Δx)²

K₂ = The final kinetic energy of the stone and the catapult system = 1/2·m·v²

Where;

v = The velocity with which the stone leaves the catapult

U₂ = The final potential energy of the stone and the catapult system = 0 J (The rubber of the catapult returns to the relaxed state)

Therefore, by substitution of the above values and equivalent expressions, we have;

0 J + 1/2·k·(Δx)² = 1/2·m·v² + 0 J

1/2·k·(Δx)² = 1/2·m·v²

We substitute the given known values as follows;

1/2 × 500 N/m × (0.04 m)² = 1/2 × 0.05 kg × v²

∴ v² = (1/2 × 500 N/m × (0.04 m)²)/(1/2 × 0.05 kg) = 16 m²/s²

v = √(16 m²/s²) = 4 m/s

The velocity with which the stone leaves the catapult = v = 4 m/s.