Question

A 10 kg block is originally moving at a speed of 5 m/sec. It hits a relaxed spring of k= 3000 n/m. If the block loses 10% of its original energy, how far will the spring compress when the block stops?

Answer 1

Answer:

0.27 m

Explanation:

The initial energy of the block (kinetic energy) is given by:

$E_i = K=\frac{1}{2}mv^2=\frac{1}{2}(10 kg)(5 m/s)^2=125 J$

where m=10 kg is the mass of the block and v=5 m/s is the speed.

Later, the block loses 10% of its original energy, so its new energy is 90% of the original energy:

$E_f = 0.90 E_i = 0.90 (125 J)=112.5 J$

Finally, the block transfers all its kinetic energy to the spring, so the energy is converted into elastic potential energy of the spring, given by

$E=\frac{1}{2}kx^2$

where k=3000 N/m is the spring constant and x is the compression of the spring. Solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(112.5 J)}{3000 N/m}}=0.27 m$