Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 138: 63

a). $1.08m/s^{2}$ b). $0m/s^{2}$

Maximum value of static friction =$0.5\times50\times9.8=245N$ a). Since applied force is greater than maximum value of static friction, the object must be in motion. So, $a=\frac{F-coefficient\, of\, kinetic\, friction \times 50\times9.8}{50}=\frac{250-0.4\times 50\times9.8}{50}=1.08m/s^{2}$ b). Since the force applied is less than static friction force, object will be at rest. So, $ a=0m/s^{2}$