Question

A 10 kg runaway grocery cart runs into a spring with a spring constant of 270 N/m and compresses it by 62 cm.What was the speed of the cart just before it hit the spring?

Answer 1

The speed of the 10 kg runaway grocery cart just before it hit the spring was approximately 3.92 m/s.

Step-by-step explanation:

To determine the speed of the grocery cart just before hitting the spring, we can use the conservation of mechanical energy. The potential energy stored in the compressed spring is equal to the initial kinetic energy of the cart. The potential energy stored in the spring
`(\(PE_s\))` is given by the formula
`\(PE_s = (1)/(2)kx^2\)`, where
`\(k\)` is the spring constant and
`\(x\)` is the compression of the spring.

`\[PE_s = (1)/(2) * 270 \, \text{N/m} * (0.62 \, \text{m})^2 = 50.22 \, \text{J}\]`

This potential energy is equal to the initial kinetic energy of the cart
`(\(KE\))`, which is given by the formula
`\(KE = (1)/(2)mv^2\),` where \(m\) is the mass of the cart and
`\(v\)` is its velocity.

`\[50.22 \, \text{J} = (1)/(2) * 10 \, \text{kg} * v^2\]`

Solving for
`\(v\),` we find
`\(v \approx 3.92 \, \text{m/s}\)`. Therefore, the speed of the cart just before it hit the spring was approximately 3.92 m/s. This calculation assumes no energy losses to friction or other factors, making it an idealized scenario based on the conservation of mechanical energy.