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

Answer:

F = 47.01 N

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

                       

                               

Answer 2

Answer:

F = 43.9 N

Explanation:

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