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 = v^{2}- v_{0}^{2} /2s
= 1.5^{2}/2 × 10
= 0.1125 m / sec^{2}
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