A spring,with k = 2500N/m,is fixed vertically to the floor.A 10.0-kg box is pressed down on the spring such that the total compression of the spring is 0.5 m.Assuming the work done against drag (air resistance)is 60 J,how high will the box reach above its release position?

Question

Quizplus · Accepted Answer

The potential energy stored in the compressed spring is given by:
$U_s = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the compression of the spring.
Plugging in the given values, we get:
$U_s = \frac{1}{2}(2500	ext{ N/m})(0.5	ext{ m})^2 = 625	ext{ J}$
However, we are told that there was work done against air resistance, so the actual kinetic energy of the box will be less than the potential energy stored in the spring. The work done against air resistance is given as 60 J, so the initial kinetic energy of the box is:
$K_i = U_s - W_{drag} = 625	ext{ J} - 60	ext{ J} = 565	ext{ J}$
When the box reaches the highest point, it will have zero kinetic energy and maximum potential energy (at the highest point, all the energy is in the form of potential energy). Therefore, we can use conservation of energy to find the height:
$U_g = mgh$, where $U_g$ is the gravitational potential energy, $m$ is the mass of the box, $g$ is the acceleration due to gravity, and $h$ is the height above the release position.
Setting the initial total energy equal to the final total energy (since energy is conserved), we get:
$K_i + U_s = U_g$
$565	ext{ J} + 625	ext{ J} = (10	ext{ kg})(9.81	ext{ m/s}^2)h$
Solving for $h$, we get:
$h = \frac{565	ext{ J} + 625	ext{ J}}{(10	ext{ kg})(9.81	ext{ m/s}^2)} = 2.58	ext{ m}$
So the box will reach a height of 2.58 m above its release position. Therefore, the answer is choice C.

A Spring,with K = 2500N/m,is Fixed Vertically to the Floor