A 1.5-kg box that is sliding on a frictionless surface with a speed of 12 m/s approaches a horizontal spring. The spring has a spring constant of 2000 N/m. (a) How far will the spring be compressed in stopping the box? (b) How far will the spring be compressed when the box's speed is reduced to half of its initial speed?

2. Originally Posted by pvcmd1
A 1.5-kg box that is sliding on a frictionless surface with a speed of 12 m/s approaches a horizontal spring. The spring has a spring constant of 2000 N/m. (a) How far will the spring be compressed in stopping the box? (b) How far will the spring be compressed when the box's speed is reduced to half of its initial speed?
There are no non-conservative forces present, so mechanical energy is conserved. Thus
$\frac{1}{2}mv_0^2 + \frac{1}{2}kx_0^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

Initially the spring is uncompressed, so $x_0 = 0$. At the end the box is not moving (for an instant.) So $v_0 = 0~m/s$. So
$\frac{1}{2}mv_0^2 = \frac{1}{2}kx^2$

-Dan

3. Hi,

part a)

Given frictionless surface.
Mass of box, m = 1.5 kg
velocity of the box, v = 12 ms^-1
Spring constant, k = 2000 N/m

From principle of conservation of energy:
when the box is stopped by the spring, all the kinetic energy in to box will be converted to elastic potential energy inside the spring. Hence we can say that:

Kinetic energy of box = elastic potential energy of spring
$\frac{1}{2}m\times v^2 = \frac{1}{2}k\times x^2 \longrightarrow(1)$

where,
x = compression of the spring

substituting known values into equation (1)
$\frac{1}{2}\times 1.5 \times 12^2 = \frac{1}{2}\times 2000 \times x^2$
$1.5 \times 12^2 = 2000 \times x^2$
$x^2 = \frac{1}{2000} \left(1.5 \times 12^2 \right)$
$x= 0.328633534 meters = 32.8633 centimeters$

part b)

when the velocity of the box is reduced to half its speed in part a)

v= 6 ms^-1

using the same calculations as in part a)

x = 0.164316767 meters = 16.4316 centimeters

$\frac{1}{2}m\times v^2 = \frac{1}{2}k\times x^2 \longrightarrow(1)$
This is not a comment about your analysis, which is correct. However I have noted in a couple of recent posts you using the notation $a \times b$ for multiplication. Just an FYI: There could be some confusion between simple multiplication and the cross product.