Question

In the arrangement shown, by what acceleration the boy must go up so that 100 kg block remains stationary on the wedge. The wedge is fixed and friction is absent everywhere. $(takeg=10m/s2)$

• A. $2m/s^2$
• B. $4m/s^2$
• C. $6m/s^2$
• D. $8m/s^2$

Answer 1

Answer: (A)

$2m/s^2$

Suppose the boy is going up with acceleration a for man

$T-mg=ma$

for block kept on the inclined plane:

$T=mgcos{53}^0$
$=100\times 10\times \frac{3}{5}$
$T=600$

Substitute the value of Tension in the force equation for the man.

$\therefore 600-500=50\left(a\right)$

$\therefore a=2m/s^2$