Question

A block of mass 2 kg is connected by a spring of
spring constant 200 N/m and vertically hung in freely
falling cabin. The extension in the spring will be
(1) 50 cm
(2) 10 cm
(3) 20 cm
(4) Zero​

Answer 1

Answer:

kx=mg

$x=\frac{20}{200}$

=0.1 m

=10cm

(2) 10 cm

Answer 2

The Correct Answer Is Option (3) 20 Cm

GIVEN

Mass of the block = 2 kg

Spring constant = 200 N/m

TO FIND

Extension of the wire

SOLUTION

We can simply solve the above problem as follows-

Let the maximum elongation of the spring be, x meter.

We know that,

Work done = change in potential energy of the spring.

Work done = Force × distance.

Therefore,

$mg \times x = \frac{1}{2} k {x}^{2}$

Where,

m = mass of the block

g = acceleration due to gravity = 9.8m/s2

k = spring constant

x = Extension in the spring.

Putting the values in the above equation we get,

$2 \times 9.8 \times x = \frac{1}{2} \times 200 \times {x}^{2}$

$19.6x = 100 {x}^{2}$

x = 19.6/100

= 0.196 m = 19.6 cm

19.6 ~ 20 cm

Hence, the correct answer is option (3) 20 cm

#spj2