Question

The block has a mass of 0.87 kg and moves within the smooth vertical slot. It starts from rest when the attached spring is in the unstretched position at A. sB = 0.15 m. Neglect the size and mass of the pulley. Hint: The work of F can be determined by finding the difference Δl in cord lengths AC and BC and using UF=FΔl.

Answer 1

F = 43.9 N

Step-by-step explanation:

Se can see the details in the pics 1 and 2, in order to explain the question.

Answer 2

F = 47.01 N

Step-by-step explanation:

Given:-

- The mass of the block m = 0.87 kg

- The block starts from rest vi = 0 m/s

- The block speed at point B v2 = 2.5 m/s

- The stretched length sb = 0.15 m

- Spring constant k = 100 N/m

Find:-

The vertical force F acting on the block by the slot.

Solution:-

- The cord lengths at point C (Lac) and point B (Lbc) are as follows:

Lac = √(0.3^2 + 0.4^2)

= 0.5 m

Lbc = √ [(0.4-0.15)^2 + 0.3^2]

= 0.391 m

- The work done is due to three forces: Weight, Spring force and applied force.

- Apply the principle of work and energy as follows:

Ta + ΣUa-b = Tb

Where, Ta : Initial kinetic energy at point A = 0 , ( vi = 0 )

ΣUa-b = -Uw + UF - Us ... ( sum of work done on the block )

Hence,

0 + -Uw + UF - Us = 0.5*m*v2^2

UF = 0.5*m*v2^2 + Uw + Us

F.ΔL = 0.5*m*v2^2 + m*g*sb + 0.5*k*sb^2

F*(0.5-0.391) = 0.5*0.87*2.5^2 + 0.87*9.81*0.15 + 0.5*100*0.15^2

F = 5.123955 / (0.5-0.391)

F = 47.01 N