A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 171 N at an angle of 30.7 ◦ above the horizontal. The box has a mass of 38.9 kg, and µk between the box and the floor is 0.19. Find the acceleration of the box.

Answer:-2.34 ms⁻²Step-by-step explanation:As per the question,We have been provided the information that is  Force at an angle 30.7° = 171 NMass of the box = 38.9 kgCoefficient of friction between box and the floor = 0.19Now,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 forceAs we know that,Frictional force is given by:f = μNWhere,μ = coefficient of frictionN = normal reactionFrom equation (i)N = mg - 171 sin 30.7°∴ f = μ ( mg - 171 sin 30.7°)     ............equation (iii) From equation (ii) and (iii), we getma + 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 = 55.861a = -2.34 ms⁻²Hence, the acceleration of the box is = - 2.34 ms⁻²