Question

02. 3/104 from the book (just part b). The system shown in the figure below has a spring two masses and two massless pulleys. The pulley cords are massless and inextensible. The system is released from rest and with the spring stretched 200 mm. Determine the distances traveled by the 4 kg cart before it comes to rest if m=3 kg. Assume no mechanical interference and no friction. and state whether the distance traveled is up or down the incline. Hint: Note that you will need to use the fact that the distance moved by the 4 kg mass and the distance moved by the m = 3 kg mass are related. First derive this relation and then use. Atte

I know the answer is 0.425 m, but I don't know how to get there. Also if possible to solve by using Energy that method would be preferred.

Answer 1

From the figure initially force on 4kg is :

Fa= mgsin(25°) + 5ku- T

F4 = 4 x 9.8 x sin(25°) + = x 150 x (0.2)2 – T

F4 = 19.57-1

Force on 3kg block is:

F3 = 27 - mg = 27 – 3 x 9.8 = 27 - 29.4

Let suppose net force on 4kg is zero so T = 19.57 N

Due to this net force on 3kg block will be positive means we can say that after release from 4 kg block will move downward and 3 kg block will move in upward direction.

Let, 4kg cart move downward on inclined plane by x distance then by the constraint we can say that 3 kg block will move upward by x/2 distance.

So,

Gain in Gravitational potential energy in 3kg is:

P.Ez = mgh = 3 x 9.8 x 2/2 = 14.7.0

Loss in Gravitational potential energy in 4kg is:

P.E4 = m4gh4 = mgxsin (25) = 4 x 9.8 x zsin(25°) = 16.57.0

As per question initially the spring is stretched by 0.2m means it is 0.2 m away from the mean position So, when it is start compressed then the compression in spring is given as:

x_c = x - 0.2

Because distance between the initial and final position it x and the distance of initial position from mean position is 0.2m so distance of final position from mean position is x - 0.2

So,

Gain in Spring potential energy is given as:

P.Ej= k(e – 0.2)2 => x 150 x (x – 0.2)2 = 75 x (x – 0.2)2

P.Efs = 75 x (1 – 0.2)2 = 75.rº – 30.0 + 3

Loss in Spring potential energy is given as:

P. Ejs = *(0.2)2 = * 150 (0.2)2 = 3J

By the energy conservation :

Loss in potential = Gain in potential energy

(16.57.0) + (3) = 14.7.+ (75.22 – 30x + 3)

75,2 – 31.87.x = 0

75.1 = 31.87

I= 0.4249m