Question

A car of mass 1000kg moving with a speed 18mk/h on a horizontal road collides with a horizontally mounted spring of spring constant 6.25×10^3N/m
What is the maximum compression of the spring?

Answer 1

Answer:

Given:

Speed of car = 18 km/hr

Mass = 1000 kg

Spring constant = 6.25 × 10³ N/m

To find:

Maximum Compression of the spring

Concept:

Kinetic energy of the car will be converted fully in the potential energy of the spring.

Conversion:

18 km/hr = 5 m/s

Calculation:

∴ KE(car) = PE(spring)

=> ½mv² = ½kx²

=> x² = mv²/k

=> x² = 1000 (5)²/{6.25 × 10³}

=> x² = 25 /(6.25)

=> x² = 4

=> x = 2 metres.

So final answer :

$\boxed{max \: compression = 2 \: metres}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}}$

• Mass (m) = 1000 kg
• Velocity (v) = 18 km/h = 5 m/s
• k = 6.25 * 10^3 N

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

Here Kinetic Energy will be equal to energy of spring,

$\large \star {\boxed{\sf{\dfrac{1}{2} mv^2 \: = \: \dfrac{1}{2}kx^2}}} \\ \\ \implies {\sf{\cancel{\dfrac{1}{2}}mv^2 \: = \: \cancel{\dfrac{1}{2}}kx^2}} \\ \\ \implies {\sf{mv^2 \: = \: kx^2}} \\ \\ \implies {\sf{x^2 \: = \: \dfrac{mv^2}{k}}} \\ \\ \implies {\sf{x^2 \: = \: \dfrac{\cancel{(10)^3} (5)^2}{6.25 \: \times \: \cancel{10^3}}}} \\ \\ \implies {\sf{x^2 \: = \: \dfrac{25}{6.25}}} \\ \\ \implies {\sf{x^2 \: = \: 4}} \\ \\ \implies {\sf{x \: = \: \sqrt{4}}} \\ \\ \implies {\sf{x \: = \: 2}} \\ \\ {\boxed{\sf{Compression \: (x) \: = \: 2 \: m}}}$