A 5.8kg box is on a frictionless 45∘ slope and is connected via a massless string over a massless, frictionless pulley to a hanging 2.1kg weight.What is the tension in the string if the 5.8kg box is held in place, so that it cannot move?What is the tension in the string once the box begins to move?

Answer:a) T = 20.601 Nb) T = 9.4585 NStep-by-step explanation:The tension in the string if the 5.8 kg box is held in place, so that it cannot move can be obtained as followsFor m = 2.1 Kg∑ Fy = 0T - W = 0  ⇒  T = W = m*g⇒  T = 2.1 Kg*9.81 m/s² ⇒  T = 20.601 N  (↑)The tension in the string once the box begins to move can be obtained as followsFor M = 5.8 Kg∑ Fx' = M*awhere x' is the direction of the slopethen∑ Fx' = M*a   ⇒  T - M*g*Sin ∅ = M*a    (I)For m = 2.1 Kg∑ Fy = m*a⇒  T - W = m*a⇒  T - m*g = m*a    (II)If we solve the system of equations that comprises I and II we will know T:T = M*m*g*(Sin ∅ -1) / (m - M)⇒  T = (5.8 Kg)(2.1 Kg)(9.81 m/s²)(Sin 45º - 1) / (2.1 Kg - 5.8 Kg)⇒  T = 9.4585 N