Answer:
-2.34 ms⁻²
Step-by-step explanation:
As per the question,
We have been provided the information that is
Force at an angle 30.7° = 171 N
Mass of the box = 38.9 kg
Coefficient of friction between box and the floor = 0.19
Now,
By drawing the free body diagram and taking the horizontal and vertical component for force 171 N(as show below)
`At equilibrium:
N + 171 sin 30.7° = mg ...................equation (i)
ma + 171 cos 30.7° = f ..................equation (ii)
Where,
f = frictional force
As we know that,
Frictional force is given by:
f = μN
Where,
μ = coefficient of friction
N = normal reaction
From equation (i)
N = mg - 171 sin 30.7°
∴ f = μ ( mg - 171 sin 30.7°) ............equation (iii)
From equation (ii) and (iii), we get
ma + 171 cos 30.7° = μ ( mg - 171 sin 30.7°)
Put the value of mass, μ and g = 9.8 m/s²
(38.9)a + 171 cos 30.7° = 0.19 × ( (38.9 × 9.8) - 171 sin 30.7°)
(38.9)a + 146.889 = 55.861
a = -2.34 ms⁻²
Hence, the acceleration of the box is = - 2.34 ms⁻²
