Question

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

spring which compresses 1.25 cm per 10kN. The maximum compression of the

spring is​

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:

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

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Answer 2

The maximum compression of the spring is 0.277 m.

The kinetic energy of the truck is equal to the energy required to compress the spring.

To calculate the maximum compression of the spring, we use the formula below.

⇒ Formula:

• mv²/2 = ke²/2

• mv² = ke²................ Equation 1

⇒ Where:

• m = mass of the truck
• v = velocity of the truck
• k = force constant of the spring

• e =  Maximum compression of the spring

From the question,

⇒ Given:

• m = W/g = (150×1000)/9.8 = 15306.122 kg
• v = 2 m/s
• k = (10×1000/1.25×10⁻²) = 800000 N/m
• e = ?

⇒ Substitute these values into equation 1

• 15306.122(2²) = 800000(e²)

⇒ Solve for e

• e² = (15306.122×4)/800000
• e² = 0.0765
• e = √(0.0765)
• e = 0.277 m

Hence, the maximum compression of the spring is 0.277 m.

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