A mass m1 = 3.5 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 4.8 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.71 m.

1) How much work is done by gravity on the two block system?
34 J (All of my answers are correct up to the last question, which I need help on)

2) 2) How much work is done by the normal force on m1?
0 J

3) What is the final speed of the two blocks?
2.84 m/s

4) How much work is done by tension on m1?
14.1 J

5) What is the tension in the string as the block falls?
19.9 N

7) What is the NET work done on m2?
**** This is the question I need help on. I gave the answers to the previous questions because something in there might be needed for solving this part of the problem.

1) 33.4 J
In the described system, the only available energy is the potential gravitational energy for mass m2. So: 0.71 m * 9.8 m/s^2 * 4.8 kg = 33.3984 kg*m^2/s^2 = 33.3984 J
2) 0 J
Since the table is frictionless and the normal force is towards the table's surface, there is no movement or work done via the normal force on m1. So 0 joules of work is done.
3) 2.84 m/s
Using the equation E = 0.5MV^2 and the knowledge that E = 33.3984 J from 1 above, we get
33.3984 kg*m^2/s^2 = 0.5*(3.5 kg + 4.8 kg)V^2
33.3984 kg*m^2/s^2 = 0.5*8.3 kg V^2
33.3984 kg*m^2/s^2 = 4.15 kg V^2
8.047807229 m^2/s^2 = V^2
2.83686574 m/s = V
4) 14.1 J
Using the equation E = 0.5MV^2 and the known velocity, we get
E = 0.5MV^2
E = 0.5*3.5kg* (2.84 m/s)^2
E = 0.5*3.5kg* 8.0656 m^2/s^2
E = 14.1148 kg*m^2/s^2
E = 14.1 J
5) 19.9 N
The tension in the string is the force needed to perform 14.1 J over a distance of 0.71 m. So:
14.1 J / 0.71 m = 19.88 J/m = 19.88 kg*m/s^2 = 19.88 N
7) 19.3 J
The NET work is the GROSS work performed on m2 minus the work that m2 performed. The gross work is given by 1 above (33.4 J) minus the work performed on m1 given by 4 above (14.1 J). So the net work is 33.4 - 14.1 =
19.3 J
You can also calculate the net work performed on m2 using the equation E = 0.5MV^2, so
E = 0.5MV^2
E = 0.5*4.8 kg * (2.84 m/s)^2
E = 0.5*4.8 kg * 8.0656 m^2/s^2
E = 19.35744 kg*m^2/s^2
E = 19.4 J
And as you can see, the kinetic energy that m2 has matches the NET work performed on m2, which makes sense since energy can be neither created, nor destroyed.

lhmarianateixeira lhmarianateixeira · Answer 2 · 2025-04-05T14:57:30-04:00

In this exercise we have to use the block knowledge in a system, thus we find that:

1) 33.4 J

2) 0J

3) 2.84 m/s

4) 14.1 J

5) 19.9 N

6) 19.3 J

So with the knowledge of force and energy we can calculate what is required, like this:

1) We have to calculate the value of work in the system which will be:

[tex]W=FS\\W=0.71 m * 9.8 m/s^2 * 4.8kg\\ = 33.3984 kg*m^2/s^2 \\= 33.3984 J \\=33.4 J[/tex]

2) As we are dealing with the work of the normal force, we will have to be zero, since the force will not have a distance, so:

[tex]W=FS\\W=35*0\\W=0J[/tex]

3) We want to calculate the final speed of the system as a whole, so:

[tex]E = 0.5MV^2 \\E = 33.3984 J\\ 33.3984 = 0.5*(3.5 + 4.8 )V^2\\ 33.3984 = 0.5*8.3 *V^2\\ 33.3984 = 4.15*V^2\\ 8.047807229 = V^2\\V= 2.83686574 m/s\\V=2.84 m/s[/tex]

4) We want to find the work of the voltage in this way we have:

[tex]E = 0.5MV^2\\ E = 0.5*3.5kg* (2.84 m/s)^2\\ E = 0.5*3.5kg* 8.0656 m^2/s^2\\ E = 14.1148 kg*m^2/s^2\\ E = 14.1 J[/tex]

5) Now calculating the value of the string tension, we have:

[tex]T= 14.1 J / 0.71 m \\= 19.88 J/m\\= 19.88 kg*m/s^2\\ = 19.88 N[/tex]

6) Calculating the value of work on top of block 2 we have:

[tex]E = 0.5MV^2\\ E = 0.5*4.8 kg * (2.84 m/s)^2\\ E = 0.5*4.8 kg * 8.0656 m^2/s^2\\ E = 19.35744 kg*m^2/s^2\\ E = 19.4 J[/tex]

See more about work at brainly.com/question/756198