Question

The position of a 2 kg object is given as x(t) = Bt2 +5, where x is in meters and t is in seconds. (a) Determine the force F responsible for this motion. (3) (b) If the force only results in a change in the kinetic energy of the object of 200 J between tı = 0 s and t2 = 5 s, determine the value of B. (3) (C) Without using kinematics (equations of motion), determine the displacement of the object during these 5 seconds referred to in part (b). (2)​

Answer 1

(a) Given the position function

x(t) = (B m/s²) t² + 5 m

it's clear that the object accelerates at B m/s² (differentiate x(t) twice with respect to t), so that the force exerted on the object is

F(t) = (2 kg) (B m/s²) = 2B N

(b) Recall the work-energy theorem: the total work performed on an object is equal to the change in the object's kinetic energy. The object is displaced by

∆x = x(5 s) - x(0 s)

∆x = ((B m/s²) (5 s)² + 5 m) - ((B m/s²) (0 s)² + 5 m)

∆x = 25B m

Then the work W performed by F (provided there are no other forces acting in the direction of the object's motion) is

W = (2B N) (25B m) = 50B² J = 200 J

Solve for B :

50B² = 200

B² = 4

B = ± √4 = ± 2

Since the change in kinetic energy and hence work performed by F is positive, the sign of B must also be positive, so B = 2 and the object accelerates at 2 m/s².

(c) We found in part (b) that the object is displaced 25B m, and with B = 2 that comes out to ∆x = 50 m.