A light spring of length 20 cm and force constant 2 kg
is placed vertically on a table. A small block of
mass 1 kg falls on it. The length h from the surface of
the table at which the ball will have the maximum
velocity is -
(1) 20 cm
(2) 15 cm
(3) 10 cm
(4) 5 cm
Answers
The length h from the surface of the table at which the ball will have the maximum velocity is option (2): 15 cm.
Explanation:
Given data:
The length of the light spring, l = 20 cm
Force constant, k = 2 kg/cm
The mass of the small block, m = 1 kg
To find:
The length “h” from the surface of the table when velocity is maximum
Solution:
Let the spring be compressed by a distance of “x” at a certain moment(as shown in the figure attached below).
Therefore, by using the law of conservation of energy at this moment, we can write
mg (h+x) = ½ mv² + ½ kx²
⇒ ½ mv² = mg (h+x) – ½ kx² …. (i)
[The velocity of the block will be maximum when the kinetic energy is maximum. Also, the velocity keeps on increasing as long its acceleration is increasing]
∴ d[1/2 mv²]/dx = mg – kx = 0
⇒ mg – kx = 0
⇒ x = mg/k …. (i)
Now, to get the height “h” from the surface of the table at which the velocity is maximum, we need to subtract the distance of compression in the spring from the length of the spring, so,
h = l – mg/k ……. [from (i)]
⇒ h = 20 – [1 * 9.8 / 2] ...... [taking g = 9.8 m/s²]
⇒ h = 20 – 4.9
⇒ h = 15.1 cm ≈ 15 cm
Hope this is helpful!!!!
