Question

Your friend provides a solution to the following problem. Evaluate her solution. The problem: Jim (mass 50 kg) steps off a ledge that is 2.0 m above a platform that sits on top of a relaxed spring of force constant 8000 N/m. How far will the spring compress while stopping Jim? Your friend's solution: (50kg)(9.8m/s2)(2.0m)=(1/2)(8000N/m)x x=0.25m

Answer 1

Answer:

The compress of the spring is 0.495.

Explanation:

Given that,

Mass = 50 kg

Spring constant = 8000 N/m

height = 2.0 m

We need to calculate the compress of the spring

Using law of conservation of energy

$mgh=\dfrac{1}{2}kx^2$

Where, m = mass

g = acceleration due to gravity

h = height

k = spring constant

x = distance

Put the value into the formula

$50\times9.8\times2.0=\dfrac{1}{2}\times8000\times x^2$

$x=\sqrt{\dfrac{2\times50\times9.8\times2.0}{8000}}$

$x=0.495\ m$

Hence, The compress of the spring is 0.495.

Answer 2

Answer:

$\Delta x=61.25\ mm$

Explanation:

Given:

• height of the ledge placed on a spring, $h=2\ m$

• stiffness of the spring, $k=8000\ N.m^{-1}$

• mass of the body placed over the ledge, $m=50\ kg$

Now the load on the spring due to body weight:

$w=m.g$

$w=50\times 9.8$

$w=490\ N$

As we know:

$F=k.\Delta x$

where F is the force of compression

$\Delta x=\frac{w}{k}$

$\Delta x=\frac{490}{8000}$

$\Delta x=0.06125\ m$

$\Delta x=61.25\ mm$