An object of mass 500g, initially at rest acted upon by a variable force where X component varies with X in the manner shown. The velocities of the object at point X = 8 m and X =12 m, would be the respective values of (nearly)
 

According to the work-energy theorem, the change in kinetic energy (ΔK) is equal to the work done, which is represented by the area under the force-displacement (F-x) graph.

Calculation from x = 0 to x = 8 m

The kinetic energy change is given by the equation: 
ΔK = work = area under F-x graph

For the interval from x = 0 to x = 8 m:

½ mv2 = 100 + 30 
Simplifying, we get: 
v2 = 520 
Taking the square root: 
v ≈ 23 m/s

Calculation from x = 0 to x = 12 m

For the interval from x = 0 to x = 12 m:

½ mv2 = 100 + 30 - 45 + 20 
Simplifying, we get: 
v2 = 420 
Taking the square root: 
v ≈ 20.6 m/s

Conclusion

Therefore, the appropriate values for velocity are approximately:

• 23 m/s for x = 0 to x = 8 m
• 20.6 m/s for x = 0 to x = 12 m