A5.8kg 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?

a) T = 20.601 N

b) T = 9.4585 N

Step-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 follows

For m = 2.1 Kg

∑ Fy = 0

T - 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 follows

For M = 5.8 Kg

∑ Fx' = M*a

where x' is the direction of the slope

then

∑ 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

can i get the link to the assignment?

step-by-step explanation:

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1