Smallest value of constant horizontal force:

A particle containing a mass of 8 kg starts from rest and attains a speed of 1.5 m/sec in a horizontal distance of 10 m. imagine a coefficient of friction of 0.2 and uniformly accelerated motion, what is the smallest value of any constant horizontal force P which can be needed to get this motion.

Solution

To find out acceleration of the body we have, initial velocity = 0, final velocity = 1.5 m/sec and distance travelled, s = 10 m.

∴ v ²  = v_o ²  + 2as

∴ a = v²- v₀² /2s

= 1.5²/2 × 10

= 0.1125 m / sec²

Inertia force

F_i = - m a = (-) 8 × 0.1125

    = (-) 0.9 N

Considering equilibrium of the forces, we obtain :

∑ F_x  = 0            ∴ P = F + F_i

∑ F_y  = 0  ∴ N = = 8 × 9.81 = 78.48 N

And F = 0.20 × N = 15.70 N

 Substituting these values in Eq. (1) above, we have

P = 15.70 + 0.9

= 16.60 N