$A$ block of mass $15 \;kg$ is placed on a long trolley. The coefficient of static friction between the block and the trolley is $0.18$. The trolley accelerates from rest with $0.5 \;m s^{-2}$ for $20 \;s$ and then moves with uniform velocity. Discuss the motion of the block as viewed by
$(a)$ a stationary observer on the ground,
$(b)$ an observer moving with the trolley.

(A) Mass of the block,$m = 15 \; kg$.
Coefficient of static friction,$\mu = 0.18$.
Acceleration of the trolley,$a = 0.5 \; m s^{-2}$.
According to Newton's second law of motion,the force $F$ required to move the block with the trolley is $F = ma = 15 \times 0.5 = 7.5 \; N$.
The maximum force of static friction available is $f_{max} = \mu mg = 0.18 \times 15 \times 9.8 = 26.46 \; N$ (taking $g = 9.8 \; m s^{-2}$). Even if we take $g = 10 \; m s^{-2}$,$f_{max} = 27 \; N$.
Since the required force $(7.5 \; N)$ is less than the maximum static friction $(27 \; N)$,the block will not slide on the trolley.
$(a)$ For a stationary observer on the ground,the block accelerates along with the trolley at $0.5 \; m s^{-2}$ for the first $20 \; s$ and then moves with a uniform velocity of $v = at = 0.5 \times 20 = 10 \; m s^{-1}$.
$(b)$ For an observer moving with the trolley,the block remains at rest because the frictional force provides the necessary acceleration to keep it stationary relative to the trolley.