What is the kinetic energy of the block when it is moved 2cm.

[Shivam]

Homework Statement

Given that M=20kg, the force constant k=10kN/m, the frictional force is 80N and the system is held at rest with l=l°+10cm and then released, l° is the unstreched length of the spring.then
a) What is the kinetic energy of the block when it is moved 2cm from its point of release.

Relevant Equations

Work done by all the forces on a particle = change in its kinetic energy.

My attempt at the solution...
I only have problem in solving part a)

1) i calculated the spring force actingbon the table block and it is greater than the frictionnand tension force acting in the opposite direction, so by that the block will move in left.
2) now i found that which forces are acting on the particle which are spring force, tension, and friction
3) I calculated work done by friction = -1.6N, work done by tension= -8N
And final spring potential energy=32N.mand intial spring potential energy= 50N.m
4) i equated all these values to the equation-

Wf+Wt=∆k+∆U

{Where Wf= work done by friction
Wt= work done by tension
∆U= change in potential energy due to spring
∆k= change in kenetic energy of block on table}

Now the only thing that is I can't understand that
In kinetic energy it took it = 1/2MV² where as you can see in the posted image at the lastline they took it equal to= 1/2(M+2M)V², why they took both mass when we are calculating work done on Block on the table, the work done by hanging blocknis delived through tension force.

 

[kuruman]

The solution considers a system consisting of both blocks. Then the mass of the "system" is the sum of the masses of the two blocks. You propose to use only the sliding block as your system. That's OK, but you need to understand that the tension is not constant because the block is accelerating and in fact undergoes damped harmonic motion of sorts. So the work done by the tension is a rather complicated proposition. The advantage of using both blocks as the system is that the tension becomes an internal force that does zero net work and you don't have to worry about it.

Also, what do you mean by this

3) I calculated work done by friction = -1.6N, work done by tension= -8N
And final spring potential energy=32N.mand initial spring potential energy= 50N.m

Work has units of Joules, not Newtons and so does potential energy. The unit N$\cdot$m is dimensionally correct for energy but is normally reserved for torque. I know that's what your solution sheet uses but I respectfully disagree with it.

 

[Shivam]

The solution considers a system consisting of both blocks. Then the mass of the "system" is the sum of the masses of the two blocks. You propose to use only the sliding block as your system. That's OK, but you need to understand that the tension is not constant because the block is accelerating and in fact undergoes damped harmonic motion of sorts. So the work done by the tension is a rather complicated proposition. The advantage of using both blocks as the system is that the tension becomes an internal force that does zero net work and you don't have to worry about it.

Also, what do you mean by this

Work has units of Joules, not Newtons and so does potential energy. The unit N$\cdot$m is dimensionally correct for energy but is normally reserved for torque. I know that's what your solution sheet uses but I respectfully disagree with it.

First i am sorry for writing wrong Units and thanks for the torque units and also for telling me about the tension will be variable.. can you lastly tell me how tension will have zero work as internally, if you explainna little bit..

 

[kuruman]

First i am sorry for writing wrong Units and thanks for the torque units and also for telling me about the tension will be variable.. can you lastly tell me how tension will have zero work as internally, if you explainna little bit..

It's really simple. For an ideal pulley the tension acting on each block will have the same magnitude $T$ at all times. If the hanging block moves up by $ds$ the tension does positive work $T~ds$ on it because the tension and the displacement are in the same direction. Meanwhile the other block moves to the left by $ds$ which is in a direction opposite to the tension. Therefore the tension does negative work $-T~ds$ on the sliding block. Thus, the net work done by tension on the two-block system is zero. This result is a direct consequence of Newton's 3rd law.