A 2200 g mass starts from rest and slides a distance ℓ down a frictionless 27◦ incline, where it contacts an unstressed 120 cm long spring of negligible mass as shown in the figure. The mass slides an additional 17 cm as it is brought momentarily to rest by compressing the spring of force constant 21 N/cm. The acceleration of gravity is 9.8 m/s^2. Note: The spring lies along the surface of the ramp (see figure). Assume: The ramp is frictionless. Now, the external force is rapidly removed so that the compressed spring can push up
the mass.
Find the initial separation ℓ between mass and spring. Answer in units of m.
Please help with the correct answer and a good explanation.
Physics HW help???
回答 (1)
2017-12-31 1:14 pm
Lost in gravitation potential energy of the mass
= mgh
= (2200/1000 kg) × (9.8 m/s²) × {[ℓ + (17/100)] cos27°] m}
= 2.2 × 9.8 × (ℓ + 0.17) × cos27° J
Gain in elastic potential energy of the spring
= (1/2)kx²
= (1/2) × (21 × 100 N/m) × (17/100 m)²
= 0.5 × 2100 × 0.17² J
Lost in gravitation potential energy of the mass = Gain in elastic potential energy of the spring
2.2 × 9.8 × (ℓ + 0.17) × cos27° = 0.5 × 2100 × 0.17²
ℓ + 0.17 = 0.5 × 2100 × 0.17² / (2.2 × 9.8 × cos27°)
ℓ = [0.5 × 2100 × 0.17² / (2.2 × 9.8 × cos27°)] - 0.17
ℓ = 1.41 m = 141 cm
= mgh
= (2200/1000 kg) × (9.8 m/s²) × {[ℓ + (17/100)] cos27°] m}
= 2.2 × 9.8 × (ℓ + 0.17) × cos27° J
Gain in elastic potential energy of the spring
= (1/2)kx²
= (1/2) × (21 × 100 N/m) × (17/100 m)²
= 0.5 × 2100 × 0.17² J
Lost in gravitation potential energy of the mass = Gain in elastic potential energy of the spring
2.2 × 9.8 × (ℓ + 0.17) × cos27° = 0.5 × 2100 × 0.17²
ℓ + 0.17 = 0.5 × 2100 × 0.17² / (2.2 × 9.8 × cos27°)
ℓ = [0.5 × 2100 × 0.17² / (2.2 × 9.8 × cos27°)] - 0.17
ℓ = 1.41 m = 141 cm
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