A5.8 kg box is on a frictionless 37 ∘ slope and is connected via a massless string over a massless, frictionless pulley to a hanging 1.7 kg weight. what is the tension in the string once the box begins to move?

T=20.62 N

Step-by-step explanation:

Given

inclination

Let T be the tension

From diagram

---1

For block

--2

from  1 & 2 we get

(a). The tension in the string if the 4 kg box held in place is 22.48 N.

(b). The 4 kg box moves on downward.

(c). The tension in the string once the box begins to move is 24.4 N.

Explanation:

Given that,

Mass of box m₁ =4 kg

Mass of second box m₂= 2 kg

Angle = 35°

(a). We need to calculate the tension in the string if the 4 kg box is held in place, so that it cannot move

Using formula of tension

Put the value into the formula

(b). If the box is then released, which way will it move on the slope

We need to calculate the tension for block of mass 2 kg

put the value into the formula

Here, T > T'

So, the first block moves on downward.

(c). We need to calculate the acceleration

Using formula of acceleration

Put the value into the formula

We need to calculate the tension in the string once the box begins to move

For mass 4 kg

Using balance equation

Put the value into the formula

Hence, (a). The tension in the string if the 4 kg box held in place is 22.48 N.

(b). The 4 kg box moves on downward.

(c). The tension in the string once the box begins to move is 24.4 N.

(a) 19.62 N

(b) Box moves down the slope

(c) 24.43 N

Explanation:

(a)  

2 Kg box  causes tension

hence where m is mass and g is gravitational force  

T'=4*9.81 sin 35= 22.5071 N  

Since T' is greater than , then the box moves down the slope  

(c)  

Acceleration a=  

When moving, the box will exert force T"=  

T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N

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

correct answer is option b.

step-by-step explanation:

the true statement about geometric figure is option b that is one line can be drawn through all three points.

if three points are on the same plane, they can be connected by a line.