Question

53.
a 4 kg object at a height of 1 m is pushed with a velocity of 5 m/s towards a spring, as shown in the figure. if the surface is frictionless and the spring constant is 200n/m, what is the maximum compression of the spring?

Explanation:

Step1: Apply Conservation of Energy

The total mechanical energy (kinetic + potential) at the initial position equals the elastic potential energy at maximum compression (where kinetic energy is zero). The initial energy is the sum of gravitational potential energy ($mgh$) and initial kinetic energy ($\frac{1}{2}mv^2$). The final energy is the elastic potential energy of the spring ($\frac{1}{2}kx^2$), where $x$ is the compression. So:
$$mgh + \frac{1}{2}mv^2 = \frac{1}{2}kx^2$$

Step2: Substitute Values

Given $m = 4\ \text{kg}$, $g = 9.8\ \text{m/s}^2$, $h = 1\ \text{m}$, $v = 5\ \text{m/s}$, $k = 200\ \text{N/m}$. Substitute into the equation:
Left - hand side (LHS):
$mgh = 4 \times 9.8 \times 1 = 39.2\ \text{J}$
$\frac{1}{2}mv^2 = \frac{1}{2} \times 4 \times 5^2 = 50\ \text{J}$
Total LHS: $39.2 + 50 = 89.2\ \text{J}$

Right - hand side (RHS): $\frac{1}{2} \times 200 \times x^2 = 100x^2$

Step3: Solve for $x$

Set LHS = RHS:
$$100x^2 = 89.2$$
$$x^2 = \frac{89.2}{100} = 0.892$$
$$x = \sqrt{0.892} \approx 0.944\ \text{m}$$ (or using $g = 10$ for approximation: LHS = $4\times10\times1+\frac{1}{2}\times4\times25 = 40 + 50 = 90$, $x^2=\frac{90}{100}=0.9$, $x=\sqrt{0.9}\approx0.948\ \text{m}$, close to the precise value)

Answer:

The maximum compression of the spring is approximately $\boldsymbol{0.94\ \text{m}}$ (or $\approx 0.95\ \text{m}$ if using $g = 10$ for simplicity).