Question

plz send me the correct answer​

Answer 1

$\huge\mathfrak\green{☟ \: \: \: answer \: \: \: \: ✎}$

Let T be the tension in the string and a be the common acceleration in the direction shown in Fig.

The equations of motion of two blocks are :

$50g \: sin \: 30° - T = 50a$

$T - 30g \: sin \: 60° = 30a$

$Adding \: (i) \: and \: (ii) , \: we \: get$

$50g \: sin \: 30° - 30g \: sin \: 60° = 80a$

$50 \times 10 \times \frac{1}{2} - 30 \times 10 \frac{ \sqrt{3} }{2} = 80a$

$a = \frac{250 - 150 \times 1.732}{80} = \frac{ - 9.8}{80} = - 0$

$1225/{s}^{2}$

Negative Sign indicates that 50 kg block will slide up the plane instead of sliding down

From

$(i) ,T = 50g \: sin \: 30° - 50a \: T = 50 \times 10 \times \frac{1}{2} - 50( - 0.1225)$

$= 256.12n$

$\\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸} \\ and\\ \sf \colorbox{lightgreen} {\red❤ANSWER ᵇʸ ᶠˡⁱʳᵗʸ ᵇᵒʸ}$

Answer 2

Answer :-

• The tension in the string = 256 N
• Acceleration of the two blocks = 0.1225

Explanation :-

• Now first of all we find FBD of this diagram
• { for the first block }
• Mass of first block = 50 kg
• angle = 30°
• g = gravitational force = 9.8 m/s² = 10 m/s² ( nearly )
• a = ?
• T = ?

• hence , FBD of first block
• mg sin∅ - T = m a
• 50 × g sin∅ - T = 50 × a
• 50 g sin 30° - T = 50 a _____( 1 )

• Now for the second block
• Mass of the second block = 30
• angle = 60°
• g = gravitational force = 9.8 m/s² = 10 m/s² ( nearly )
• a = ?
• T = ?
• T - m g sin ∅ = ma
• T - 30 g sin 60° = 30 a _____( 2 )

• Now eliminate to these equations no. (1) and (2)
• hence we get =
• 50 g sin 30° - 30 g sin 60° = 80 a

$50 \: g \times \frac{1}{2} - 30g \times \frac{ \sqrt{3} }{2} = 80a \\ \\ 25 \: g - 15 \sqrt{3} g = 80a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ 25g - 25.98g = 80a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ 0.98g = 80a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\0.98 \times 10 = 80a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ a = \frac{0.98 \times 10}{80} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\{ \pink{ a = 0.1225 }}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now for tension :-

• we put the value of acceleration in eq. (1) or (2)
• hence

$50g \times \sin30 \degree - ᴛ = \: - \: 50a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ 50 \times 10 \times \frac{1}{2} - ᴛ = - 50 \times 0.1225 \\ \\ 250 - ᴛ = - 6.15 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ 250 + 6.15 = ᴛ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ 256.15 = ᴛ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ ᴀɴsᴡᴇʀ ɴᴇᴀʀʟʏ \: \: \: \: \: \\ \\ \pink{ᴛ = 256}$