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7) A 15 kg runaway grocery cart runs into a spring with spring constant 260 N/m and compresses it by 62 cm .

A. What was the speed of the cart just before it hit the spring?
Express your answer with the appropriate units.

User Jelly Bean
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1 Answer

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Answer: 2.53 m/s

Explanation: To solve this problem, we can use the conservation of energy principle. At the initial moment, the cart has kinetic energy only and at the final moment, it has potential energy only, stored in the compressed spring.

The initial kinetic energy of the cart can be calculated as:

KE = (1/2)mv^2

where m is the mass of the cart and v is its speed just before it hits the spring.

The final potential energy of the compressed spring can be calculated as:

PE = (1/2)kx^2

where k is the spring constant and x is the compression of the spring.

According to the conservation of energy principle, the initial kinetic energy is equal to the final potential energy, so we can set these two expressions equal to each other and solve for v:

(1/2)mv^2 = (1/2)kx^2

v^2 = (k/m)x^2

v = sqrt[(k/m)x^2]

Plugging in the given values, we get:

v = sqrt[(260 N/m / 15 kg) * (0.62 m)^2] ≈ 2.53 m/s

Therefore, the speed of the cart just before it hit the spring was approximately 2.53 m/s.

User Salizar Marxx
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