Final answer:
The kinetic energy of the block when it is 5cm away from the equilibrium position is found using the conservation of mechanical energy and is calculated to be 0.19J, which is option C.
Step-by-step explanation:
When a block of mass 1kg is attached to a spring with a spring constant of 50Nm-1 and is displaced by 10cm from its equilibrium position, it stores potential energy in the spring. As the block moves towards the equilibrium position, this potential energy converts into kinetic energy. At a distance of 5cm from the equilibrium position, we can calculate the kinetic energy using the conservation of mechanical energy.
The potential energy of the spring when the block is 10cm away is given by P
E = (1/2)kx2
where k is the spring constant and x is the displacement.
At 10cm (0.10m) displacement, PE = (1/2) * 50N/m * (0.10m)2 = 0.25J.
When the block is 5cm (0.05m) away from equilibrium, the potential energy is
PE = (1/2) * 50N/m * (0.05m)2 = 0.0625J.
The difference in potential energy between these two positions (0.25J - 0.0625J = 0.1875J) represents the kinetic energy at 5cm from equilibrium position, since no other work is done on the system (frictionless surface).
Therefore, the kinetic energy of the block is 0.1875J, which we round to 0.19J, matching option C.