Q. 351. Dynamics

Question 351.
A force F, acts in a constant direction on a 10 kg block as shown in the figure. The magnitude of the force varies with the position s of the block, as given by F = 60s^2. When s = 0, the block is moving horizontally to the right with a speed of 3 m/s. If friction exists between the block and the ground such that the coefficient of kinetic friction is u = 0.3, what is the speed of the block at s = 2m.

Answer 351.
Horizontal component of force,
Fh = 0.8 F
Vertical component f force,
Fv = 0.6 F
Weight of the block,
W = 10 * 9.81 = 98.1 N
Normal reaction from the ground
R = 98.1 + 0.6 F
Frictional force from the ground
= μR
= (0.3) * (98.1 + 0.6 F)
Frictional force at a distance s
= (0.3) * (98.1 + 0.6 * 60s^2)
= 29.43 + 10.8 s^2
Net force at a distance s
= 0.8 F - (29.43 + 10.8 s^2)
= 48s^2 - 29.43 - 10.8 s^2N
= 37.2 s^2 - 29.43

Work done for a small displacement, ds
dW = (37.2 s^2 - 29.43) ds
Work done for displacement from s = 0 to s = 2 m
W = ∫ (s = 0 to 2) (37.2 s^2 - 29.43) ds
=> W = [12.4 s^3 - 29.43s] ... (s = 0 to 2)
=> W = 12.4 * 2^3 - 29.43 * 2
=> W = 40.34 J

Work done = change in KE
=> (1/2) * 10 (v^2 - 3^2) = 40.34
=> v^2 - 9 = 8.045
=> v^2 =17.045
=> v = 4.13 m/s.

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