Question

A 0.20kg object moves along a straight line. The net force acting on the object varies with the object's displacement as shown in the graph. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20m. Determine each of the following.

a. The acceleration of the particle when its displacement x is 6m.

b. The time taken for the object to be displaced the first 12 m. 

c. The amount of work done by the net force in displacing the object the first 12m. 

d. The speed of the object at displacement x = 12m. 

e. The final speed of the object at displacement x = 20m. 

f. The change in the momentum of the object as it is displaced from x = 12m to x = 20m

Answer 1

(a) The force is constant, so simple F_net = ma is sufficient. (4) = (0.2) a a = 20m/s²

(e) The area under the triangle will give the extra work for the last 8m 

½(8)(4) = 16J + work for first 12m (48J) = total work done over 20 m = 64J 

Again using work energy theorem 

W = ½ m vf 2 64J = ½(0.2)v_f² 

 v_f = 25.3m/s 

Note: if using F = ma and kinematics equations, the acceleration in the last 8 m would need to be found using the average force over that interval.

(f) The momentum change can simply be found with ∆p = m∆v = m(v_f – v_i) = 0.2 (25.3 – 21.9) = 0.68kgm/s