Question

A system is shown in the figure. Block A moves with velocity 10 m/s. The speed of the mass B will be
4515
10m/s
A
20
(A) 10 √2 m/s
(B) 5 1/3 m/s
m/s
(D) 10 m/s​

Answer 1

The speed of the mass B will be 20√3 m/s.

Explanation:

Let the rectangular component of velocity of mass B = Vx and Vy

• Net velocity along string BC is

Vxsin(45°)+ Vycos(45°) = 10

Vx+Vy=10√2   ....(1)  

• Net velocity along string BA is

Vxcos (75°) −Vycos(15°) =0......(2)

Solving equations (2) and (1)

V= √V^2x+V^2y

 = 20√3 m/s

Thus the speed of the mass B will be 20√3 m/s.

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