Question

On a horizontal frictionless surface there is a block with the mass of 1 kg bound to a spring with a negligible mass. Towards the end of the spring there is another block moving with the speed of 4 m/s with the mass of 4 kg. What is the maximum work on the spring if his elasticity constant is 100 N/m.

Answer 1

Step-by-step explanation:

The maximum work done on the spring can be calculated using the principle of conservation of energy. Initially, all the energy is kinetic energy of the moving block, and at the maximum compression of the spring, all this energy will be stored as potential energy in the spring.

The initial kinetic energy (KE) of the moving block can be calculated using the formula:

KE = 0.5 * m * v^2

where m is the mass of the block (4 kg) and v is its velocity (4 m/s).

So, KE = 0.5 * 4 kg * (4 m/s)^2 = 32 Joules

At maximum compression, all this kinetic energy will be converted into potential energy (PE) stored in the spring. The potential energy stored in a spring is given by:

PE = 0.5 * k * x^2

where k is the spring constant (100 N/m) and x is the displacement (or compression) of the spring. We want to find out this PE which will be equal to the initial KE.

So, 32 Joules = 0.5 * 100 N/m * x^2

Solving for x, we get x = sqrt((2 * 32 Joules) / (100 N/m)) = 0.8 m

So, the maximum work done on the spring, which is equal to the maximum potential energy stored in it, is **32 Joules**.