Question

A truck weighing 150kN and travelling at 2m/sec impacts with a buffer

Answer 1

The maximum compression of the spring is 0.276 m

Step-by-step explanation:

First of all, we calculate the spring  constant of the spring, by using Hooke's law:

$F=kx$

where

F = 10 kN = 10,000 N is the force applied

x = 1.25 cm = 0.0125 m is the corresponding compression of the spring

k is the spring constant

Solving for k,

$k=\frac{F}{x}=\frac{10,000}{0.0125}=8\cdot 10^5 N/m$

When the truck impacts with the spring, all the kinetic energy of the truck is converted into elastic potential energy of the spring, so we can write

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

where

m is the mass of the truck

v = 2 m/s is the initial speed of the truck

$k=8\cdot 10^5 N/m$ is the spring constant

x is the compression of the spring

The mass of the truck can be found from its weight:

$m=\frac{W}{g}=\frac{150,000 N}{9.81 N/kg}=15,290 kg$

So now we can re-arrange eq.(1) to find the compression, x:

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(15,290)(2)^2}{8\cdot 10^5}}=0.276 m$

Learn more about kinetic energy:

brainly.com/question/6536722

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