A block (mass = 4.0 kg) sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 100 N/m) which has its other end fixed. If the maximum distance the block slides from the equilibrium position is equal to 20 cm, what is the speed of the block at an instant when it is a distance of 16 cm from the equilibrium position?
A) 71 cm/s
B) 60 cm/s
C) 80 cm/s
D) 87 cm/s
E) 57 cm/s

Question

Quizplus · Accepted Answer

The total energy (kinetic + potential) is conserved in this system. At the equilibrium position the spring is compressed and has potential energy (PE) equal to $\frac{1}{2}kx^{2} = \frac{1}{2}(100N/m)(0.20 m)^{2} = 2 J$.
At the maximum position the block has only potential energy $PE = \frac{1}{2}kx^{2} = \frac{1}{2}(100 N/m)(0.40 m)^{2} = 8 J$. The difference $E_{d} = 6 J$ is converted into kinetic energy at this point.
At the point where the block is at a distance of 16 cm from the equilibrium position it still has some potential energy left, which is $PE = \frac{1}{2}kx^{2} = \frac{1}{2}(100 N/m)(0.04 m)^{2} = 0.08 J$. Thus, at this point the kinetic energy should be $E_{k} = E_{tot} - PE = 6 J - 0.08 J = 5.92 J$.
Using the formula $E_{k} = \frac{1}{2}mv^{2}$, we find the speed:
$v = \sqrt{\frac{2E_{k}}{m}} = \sqrt{\frac{2(5.92 J)}{4.0 kg}} \approx 6.0 m/s \approx 60 cm/s$.

A Block (Mass = 4