A frame of mass $200 g$ when suspended from a massless spring extends it by $10 c m$ . A lump of clay of mass $200 g$ is dropped from rest on to the frame from a height of $30 c m$ as shown in figure. What is the maximum distance through which pan moves downwards?

Open in App

Solution

$m g = k x$
$\Rightarrow \frac{2}{10} \times 10 = k \frac{10}{100}$
$\Rightarrow k = 20$
Now,
KE gained +PE lost $=$ PE of spring
$\Rightarrow \frac{1}{2} \times \frac{2}{10} {(\sqrt{2 \times 10 \times \frac{3}{10}})}^{2} + \frac{2}{10} \times 10 (0.3 + x) + \frac{2}{10} \times 10 \times 20$
$= \frac{1}{2} \times 20 x^{2}$
$\Rightarrow 10 x^{2} - 4 x - 1.220$
$x = 0.6 c m$
$= 60 c m$
Hence, the answer is $60 c m .$

Suggest Corrections

Similar questions

Q. A frame of mass 200 g when suspended from a masselss spring extends is by 10 cm . A lump of clay of mass 200 g is dropped from rest on to the frame through which pan moves downward?

Q. A block of mass 200 g is suspended through a vertical spring. The spring is stretched by 1.0 cm when the block is in equilibrium. A particle of mass 120 g is dropped on the block from a height of 45 cm. The particle sticks to the block after the impact. Find the maximum extension of the spring.
Figure

Q. A massless platform is kept on a light elastic spring, as shown in the figure. When particle of mass

0.1 k g

is dropped on the pan from a height of

0.24 m

, the particle strikes the is pan, and the spring is compressed by

0.01 m

. From what height should the particle be dropped to cause a compression of

0.04 m

A block of mass 200 g is suspended through a vertical spring.The spring is stretched by 1.0 cm when the block is in equilibrium.A particle of mass 120 g is dropped on the block from a height of 45 cm.The particle sticks to the block after the impact.Find the maximum extension of the spring.Take g=10 $m / s^{2}$ .

Q. A massless platform is kept on a light elastic spring, as shown in the figure. When a particle of mass

0.1 kg

is dropped on the pan from a height of

0.24 m

, the spring gets compressed by

0.01 m

. From what height should the same particle be dropped to cause a compression of

0.04 m

in the spring?
Neglect the effect of collision.

Join BYJU'S Learning Program

A frame of mass 200 g when suspended from a massless spring extends it by 10 cm. A lump of clay of mass 200 g is dropped from rest on to the frame from a height of 30 cm as shown in figure. What is the maximum distance through which pan moves downwards?

mg=kx⇒210×10=k10100⇒k=20Now, KE gained +PE lost = PE of spring ⇒12×210(√2×10×310)2+210×10(0.3+x)+210×10×20=12×20x2⇒10x2−4x−1.220 x=0.6cm =60cmHence, the answer is 60cm.

A frame of mass $200 g$ when suspended from a massless spring extends it by $10 c m$ . A lump of clay of mass $200 g$ is dropped from rest on to the frame from a height of $30 c m$ as shown in figure. What is the maximum distance through which pan moves downwards?