Answer:
The resulting acceleration of the box is 3.92 [tex]m/s^2[/tex]
Explanation:
Please refer to the Free Body Diagram
According to the second Newton's Law, the acceleration of the mass will depend on the net force applied to it.
In the y-axis, the net force is zero since the mass won't move in that direction. We only need to analyze the dynamics on the x-axis.
The problem states that initially, the box moves at a constant speed which means zero acceleration, or zero net force.
If we analyze the forces on the x-axis we find:
F - Fr = m.a
Where F is the originally applied force, Fr is the Friction force, m is the mass of the box and a is the initial acceleration, which we found to be zero. Thus:
F - Fr = 0 => F = Fr = [tex]\mu[/tex].N, being N the Normal force and [tex]\mu[/tex] the kinetic friction coefficient
By analyzing the y-axis, we find N = W = m.g
So N = 50 kg. 9.8 [tex]m/s^2[/tex] = 490 Nw
The Friction force is then:
Fr = 490 Nw . 0.2 = 98 Nw
Which gives us the initial Force:
F = 98 Nw
When tripled, the new Force will be
F' = 294 Nw
And the corresponding x-axis equilibrium condition is:
F' - Fr = m.a' ....(a' is the resulting acceleration). So we have:
[tex]a=\frac{F'-Fr}{m}=\frac{294Nw-98Nw}{50 Kg} =\frac{196Nw}{50Kg}[/tex]
a = 3.92 [tex]m/s^2[/tex]
