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

0.27 m

Step-by-step explanation:

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

`E_i = K=(1)/(2)mv^2=(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=(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{(2E)/(k)}=\sqrt{(2(112.5 J))/(3000 N/m)}=0.27 m`