Answer:
a = 2.012 m/s²
Explanation:
given,
Mass of the metal safe, M = 105 Kg
Pushing Force,F= 520 N
coefficient of kinetic friction, μ_k = 0.3
acceleration of the safe on the floor = ?
frictional force acting on the safe
[tex]f = \mu_kN[/tex]
[tex]f = \mu_k\times m g[/tex]
[tex]f = 0.3\times 105 \times 9.8[/tex]
[tex]f = 308.7 N[/tex]
now, Net Force acting on the safe
[tex]F_{net} = F - f[/tex]
[tex]F_{net} =520-308.7[/tex]
[tex]F_{net} = 211.3\ N[/tex]
we know,
F = m a
[tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{211.3}{105}[/tex]
a = 2.012 m/s²
Hence, the acceleration of the safe is equal to a = 2.012 m/s²
