Find the magnitude of acceleration of centre of mass of two blocks as shown in figure. Neglect friction everywhere. (g = 10 ms $^{2}$ )

$\frac{1}{\sqrt{2}}$ ms $^{2}$
1 ms $^{2}$
$\sqrt{2}$ ms $^{2}$
zero

Question

Find the magnitude of acceleration of centre of mass of two blocks as shown in figure- Neglect friction everywhere- -g - 10 ms-2-displaystyle frac-1-sqrt 2- ms-2-1 ms-2-sqrt 2 ms-2-zero

Toppr Admin · Accepted Answer

A-mgsin53o-x2212-mgsin37o2m-1ms-x2212-2acm-x2223-x2223-x2223-m1bara1-m2-xAF-a2m1-m2-x2223-x2223-x2223-a-x221A-2-1-x221A-2ms-x2212-2

Find the magnitude of acceleration of centre of mass of two blocks as shown in figure. Neglect friction everywhere. (g = 10 ms2)1√2 ms21 ms2√2 ms2zero

a=mgsin53o−mgsin37o2m=1ms−2acm=∣∣∣m1bara1+m2¯a2m1+m2∣∣∣=a√2=1√2ms−2

Find the magnitude of acceleration of centre of mass of two blocks as shown in figure. Neglect friction everywhere. (g = 10 ms $^{2}$ )

$\frac{1}{\sqrt{2}}$ ms $^{2}$
1 ms $^{2}$
$\sqrt{2}$ ms $^{2}$
zero

$a = \frac{m g s i n 53^{o} - m g s i n 37^{o}}{2 m} = 1 m s^{- 2}$
$a_{c m} = ∣ ∣ ∣ \frac{m_{1} b a r a_{1} + m_{2} {¯ a}_{2}}{m_{1} + m_{2}} ∣ ∣ ∣ = \frac{a}{\sqrt{2}} = \frac{1}{\sqrt{2}} m s^{- 2}$