Final answer:
The spring constant of a spring compressed by 0.30 m by a 4.0 kg box is 400 N/m, as determined by equating the box's initial kinetic energy to the potential energy stored in the spring at maximum compression.
Step-by-step explanation:
To determine the spring constant (k) of the spring that a 4.0 kg box compresses by 0.30 m, we can use the conservation of energy principle. The initial kinetic energy of the box when it starts sliding must equal the potential energy stored in the spring when the box comes to a stop.
The initial kinetic energy (KE) is given by:
KE = \(\frac{1}{2}mv^{2}\)
Substituting the given mass (m = 4.0 kg) and the initial speed (v = 3.0 m/s) into the equation, we get:
KE = \(\frac{1}{2}\times4.0\ kg\times(3.0\ m/s)^{2}\) = 18 J
The potential energy (PE) stored in the spring when compressed is given by:
PE = \(\frac{1}{2}kx^{2}\)
Where k is the spring constant and x is the compression distance (0.30 m). By setting KE equal to PE, we can solve for k:
18 J = \(\frac{1}{2}k\times(0.30\ m)^{2}\)
Simplifying and solving for k yields:
k = \(\frac{2\times18\ J}{(0.30\ m)^{2}}\) = 400 N/m
Therefore, the spring constant is 400 N/m, which corresponds to option D.