Question about Velocity, involving pulley 2x weights and friction is a factor.

[recreated]

Figure Q3 shows a connected system arrangement.

• The block of mass 400 kg is released from rest in the position shown,
• it pulls the 300 kg mass up the ramp of angle 30o.
• the coefficient of kinetic friction between the 300 kg mass and the ramp is 0.5

Determine the velocity of the 400 kg block when it has descended 6 m.



Please explain your answer if you give me one, and try and be sure it's correct or atleast tell me if you are not confident enough about it. I have been through soo many exsamples from 3 dif. books and I can not understand them enough to apply what I learn, probably doesn't help that they're all written by W.bolton :). And also can someone tell me that W.Bolton's books are extremely complicated so I can feel smart again?

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The Attempt at a Solution

 

[djeitnstine]

Homework forum policy is to show an attempt at a solution (as shown by #3 in bold font)

Always start by free body diagrams. You should already know this.

 

[Mene]

To determine the velocity of the 400 kg block after it has descended 6 m, we can use the principles of conservation of energy and Newton's laws of motion.

First, we can calculate the potential energy of the system at the starting position, where the 400 kg block is at a height of 6 m above the ground. This can be calculated using the formula PE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height. In this case, the potential energy would be PE = (400 kg)(9.8 m/s^2)(6 m) = 23,520 J.

Next, we can calculate the kinetic energy of the system at the bottom of the ramp, where the block has descended 6 m. This can be calculated using the formula KE = 1/2mv^2, where m is the mass of the block and v is the velocity. In this case, the kinetic energy would be KE = 1/2(400 kg)v^2.

According to the law of conservation of energy, the total energy of a system remains constant. Therefore, the potential energy at the starting position must be equal to the kinetic energy at the bottom of the ramp. This can be expressed as:

PE = KE
23,520 J = 1/2(400 kg)v^2

Solving for v, we get:

v = √(2*23,520 J / 400 kg) = 24.33 m/s

Therefore, the velocity of the 400 kg block when it has descended 6 m would be 24.33 m/s.

As for the coefficient of kinetic friction, it affects the motion of the 300 kg block being pulled up the ramp. It will cause a decrease in the acceleration of the block, therefore affecting the overall motion of the system. However, since we are only concerned with the velocity of the 400 kg block, the coefficient of kinetic friction does not directly impact our calculation.