Q: A block of mass 5 kg is pressed against a wall (in YZ-plane) with a force $\vec{F} = 40 \hat{i} + 3 0 \hat{j} + 10 \hat{k} $ . The vertical upward direction is taken as positive Z-axis. The coefficient of friction is 0.5. The acceleration of the block is
(a) zero
(b) (3.6\hat{j} – 4.8 \hat{k} )m/s²
(c) (3.6\hat{i} + 4.8 \hat{k} )m/s²
(d) (-4\hat{k} )m/s²
Ans: (b
Solution: 
Normal force on the block , N = 40 Newton
Resultant force on the block is
$ \displaystyle F = \sqrt{(10-50)^2 + (30)^2} = 50 N $
Acceleration of the block is
$ \displaystyle a = \frac{50 – f}{m} = \frac{50-\mu \times 40}{5} $
$\displaystyle a = \frac{50-20}{5} = 6 m/s^2 $
tanθ = 40/30 = 4/3
θ = 53°
$ \displaystyle \vec{a} = 6cos53\hat{j}-6sin53\hat{k} $
$ \displaystyle = 3.6\hat{j} – 4.8\hat{k} $