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A block is dropped onto a spring with k = 40 N/m. The block has a speed of 3.5 m/s just before it strikes the spring. If the spring compresses an amount 0.12 m before bringing the block to rest, what is the mass of the block?

1 Answer

2 votes

Final answer:

Using the conservation of energy, the mass of the block that compresses the spring 0.12 m can be found to be 0.49 kg by equating the kinetic energy before impact to the potential energy at maximum compression.

Step-by-step explanation:

To find the mass of the block dropped onto a spring with a spring constant (k) of 40 N/m, where the block has a speed of 3.5 m/s just before it strikes the spring, and the spring compresses 0.12 m before bringing the block to rest, we can use the conservation of energy principle. The kinetic energy (KE) of the block just before impact is transformed into the potential energy (PE) stored in the spring at maximum compression.

The equation for kinetic energy is KE = 0.5 × m × v^2, and the equation for the potential energy stored in a compressed spring is PE = 0.5 × k × x^2, where m is the mass of the block, v is the velocity, k is the spring constant, and x is the compression of the spring.

By setting the kinetic energy equal to the potential energy, we can solve for the mass of the block:

0.5 × m × (3.5 m/s)^2 = 0.5 × (40 N/m) × (0.12 m)^2

Solving for m gives us:

m = ((40 N/m) × (0.12 m)^2) / (3.5 m/s)^2 = 0.49 kg

Therefore, the mass of the block is 0.49 kg.

answered
User Daniel Delaney
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