Question

Two bodies of mass 4 kg and 6 kg are tied to
the ends of a massless string. The string
passes over a pulley which is frictionless
(see figure). The acceleration of the system in
terms of acceleration due to gravity (g) is :
4 kg
6 kg
(1) g/2
(2) g/5
(3) g/10
(4) g​

Answer 1

Refer to the attachment.

Refer to the attachment.Free body diagram of each body is also attached to the answer.

We are given a massless which has its both the ends tied to seprate bodies A and B of mass 4 kg and 6 kg respectively.

Since the body B has more mass than the body A hence it would accelerate towards the body B.

Let us examine the body B.

Here, The mass of the body B is 6 kg and acceleration due to gravity is acting in downwards direction.

So, The net downward force of body B will be,

⇒ Mg - T = Ma [ Since the body B is heavier than the body A, so the body B will move in downward direction with acceleration a ]

⇒ 6g - T = 6a

⇒ T = 6(g - a) ...(1)

Now, analysing the body A,

The mass of the body A is 4 kg.

Which is tied to a massless string which has a tension T = 6(g - a) [ from (1) ]

Here, The body A will have its force of weight in the opposite direction of the tension of the string so they would be subtracted, so the net force is

⇒ ma = T - mg [ because the body has less mass than body B so it would move upwards ]

⇒ 4a = 6(g - a) - 4g

⇒ 4a = 6g - 6a - 4g

⇒ 10a = 2g

⇒ a = 2g / 10

⇒ a = g / 5

Hence, The acceleration of the system is g/5

∴ Option (2) is correct.

Answer 2

GIVEN :-

• Tᴡᴏ ʙᴏᴅɪᴇs ᴏғ ᴍᴀss 4 ᴋg ᴀɴᴅ 6 ᴋg ᴀʀᴇ ᴛɪᴇᴅ ᴛᴏ ᴛʜᴇ ᴇɴᴅs ᴏғ ᴀ ᴍᴀssʟᴇss sᴛʀɪɴɢ .

• Tʜᴇ ᴘᴀssᴇs ᴏᴠᴇʀ ᴀ ᴘᴜʟʟᴇʏ ᴡʜɪᴄʜ ɪs ғʀɪᴄᴛɪᴏɴʟᴇss .

TO FIND :-

• Tʜᴇ ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ᴏғ ᴛʜᴇ sʏsᴛᴇᴍ ɪɴ ᴛᴇʀᴍs ᴏғ ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ᴅᴜᴇ ᴛᴏ ɢʀᴀᴠɪᴛʏ .

SOLUTION :-

☯︎ sᴇᴇ ᴛʜᴇ ᴀᴛᴛᴀᴄʜᴍᴇɴᴛ ᴘɪᴄᴛᴜʀᴇ, ғᴏʀ ғʀᴇᴇ ʙᴏᴅʏ ᴅɪᴀɢʀᴀᴍ .

ᴄᴀsᴇ - 1 :-

➪ ʜᴇʀᴇ, ᴛʜᴇ ᴅᴏᴡɴᴡᴀʀᴅ ғᴏʀᴄᴇ ᴀᴄᴛɪɴɢ ᴏɴ ᴛʜᴇ 6ᴋg ᴍᴀss ʙᴏᴅʏ [ᴅᴜᴇ ᴛᴏ ɪᴛ ɪs ʜᴇᴀᴠɪᴇʀ ᴛʜᴀɴ ᴛʜᴇ 4ᴋg ᴍᴀss ʙᴏᴅʏ] ᴀɴᴅ ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ɪs ᴅᴏᴡɴᴡᴀʀᴅ ᴀʟsᴏ ᴀɴᴅ ᴛᴇɴsɪᴏɴ ɪs ᴀᴄᴛɪɴɢ ᴜᴘᴡᴀʀᴅ ᴏɴ ᴛʜᴇ sᴛʀɪɴɢ .

✯ Sᴏ, Tʜᴇ ɴᴇᴛ ᴅᴏᴡɴᴡᴀʀᴅ ғᴏʀᴄᴇ ᴀᴄᴛɪɴɢ ᴏɴ 6ᴋg ᴍᴀss ʙᴏᴅʏ ɪs,

$\huge\red\checkmark$ $\bf\blue{Ma\:=\:Mg\:-\:T\:}$

⟹ 6 × a = 6 × g - T

⟹ T = 6g - 6a

⟹ T = 6 (g - a) -------(1)

ᴄᴀsᴇ - 2 :-

➪ Tʜᴇ ᴜᴘᴡᴀʀᴅ ғᴏʀᴄᴇ ᴀᴄᴛɪɴɢ ᴏɴ ᴛʜᴇ 4ᴋg ᴍᴀss ʙᴏᴅʏ [ᴅᴜᴇ ᴛᴏ ᴛʜᴇ 6ᴋg ᴍᴀss ʙᴏᴅʏ ᴘᴜʟʟᴇᴅ ᴛʜᴇ 4ᴋg ᴍᴀss ʙᴏᴅʏ] ᴀɴᴅ ᴛʜᴇ ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ɪs ᴜᴘᴡᴀʀᴅ ᴀɴᴅ ᴛᴇɴsɪᴏɴ ɪs ᴀᴄᴛɪɴɢ ᴜᴘᴡᴀʀᴅ ᴀʟsᴏ .

✯ Hᴇɴᴄᴇ, ᴛʜᴇ ɴᴇᴛ ᴜᴘᴡᴀʀᴅ ғᴏʀᴄᴇ ᴀᴄᴛɪɴɢ ᴏɴ ᴛʜᴇ 4ᴋg ᴍᴀss ʙᴏᴅʏ ɪs,

$\huge\red\checkmark$ $\bf\blue{Ma\:=\:T\:-\:Mg\:}$

➳ 4 × a = {6(g - a)} - 4 × g

➳ 4a = 6g - 6a - 4g

➳ 4a + 6a = 6g - 4g

➳ 10a = 2g

➳ a = 2g/10

➳ a = g/5

$\huge\red\therefore$ [2] Tʜᴇ ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ᴏғ ᴛʜᴇ sʏsᴛᴇᴍ ɪɴ ᴛᴇʀᴍs ᴏғ ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ᴅᴜᴇ ᴛᴏ ɢʀᴀᴠɪᴛʏ ɪs $\bf\green{\dfrac{g}{5}\:}$ .