Question

Two blocks of masses 5 kg and 7 kg are connected by uniform rope of mass 4 kg as shown in the figure. An upward force F = 200 N is applied on the system. (Take g = 10 m/s^2)

The tension at the top of the heavy rope is

If you spam/answer for the sake of points you will be reported. I repeat YOU WILL BE REPORTED.​

Answer 1

Answer:

Here is your answer. Hope it helps.

Given:

Mass of Block 1 (m₁) = 7kg

Mass of Block 2 (m₂) = 5kg

Mass of Rope (mₐ) = 4kg

Force acting upward (F) = 200N

Accelration due to gravity(g) = 10 m/s²

Tension on top of heavy rope (T₁) = ?

Solution:

Total mass of the given system (M) = m₁+m₂+mₐ = 7+5+4 = 16kg

F = m*a

a = $\dfrac{Force - Mg }{ total\:mass[M]}$

a = $\dfrac{200-160}{16}$

a = 2.5 m/s

Here they asking tension for top of heavy rope so we need to draw Free Body diagram (FBD) for 7kg block inorder to find it

F - m₁g - T₁ = m₁a (after drawing FBD)

Let us substitute the values,

200 - 7×10 - T₁ = 7a

200 - 70 - T₁ = 7×2.5

130-17.5 = T₁

T₁ = 112.5N

Hence the block would have tension of 112.5N at the top of the block

Answer 2

Answer:

Hey answer is provided above by some legend

Explanation:

Thanks for points