Man Jumps Horizontally: Work Done by Man

[Ameya Darshan]

Homework Statement

A man is standing on a cart of mass double the mass of man. mass of the man is m. initially the cart is at rest on a smooth ground. now, the man jumps w/ velocity v horizontally towards right wrt the cart. find the work done by the man during the process of jumping.

Homework Equations

W = Change in kinetic energy
Initial momentum = Final momentum

The Attempt at a Solution

I calculated the change in kinetic energy and equated it to the work done, but I'm not able to get the right answer, which is mv^2/3.

 

[TSny]

Hello and welcome to PF!
Your general approach to the problem sounds right. But we can't identify where you are making a mistake unless you show your work in detail.

 

[Ameya Darshan]

Hello and welcome to PF!
Your general approach to the problem sounds right. But we can't identify where you are making a mistake unless you show your work in detail.

This is what I did.

Conserving momentum: mv=2m(x)
therefore x = velocity of cart = v/2

Apply work energy theorem:
W = mv^2 + (2m)(v^2/4)/2 = 3mv^2/4

 

[TSny]

You did not use the fact that v is the velocity of the man wrt the cart, not the velocity of the man wrt the ground. You have three velocities to work with:
v = velocity of man wrt cart
v_c = velocity of the cart wrt the ground
v_m = velocity of the man wrt the ground.

How are these related? Which of these velocities should be used in the conservation of momentum equation?

 

[Ameya Darshan]

but the system is initially at rest, no? so wouldn't velocity will be imparted to the cart only after the man has jumped?

 

[TSny]

but the system is initially at rest, no?

Yes

so wouldn't velocity will be imparted to the cart only after the man has jumped?

Yes.

Once the man has jumped, he has a certain velocity wrt the ground (v_m) and the cart has a certain velocity wrt the ground (v_c). Also after the man has jumped, the man has a certain velocity wrt the cart (v).

 

[Ameya Darshan]

I got the right answer! Thanks!
And thank you for the welcome. :-)

 

[Anvi]

Yes
Yes.

Once the man has jumped, he has a certain velocity wrt the ground (v_m) and the cart has a certain velocity wrt the ground (v_c). Also after the man has jumped, the man has a certain velocity wrt the cart (v).

so what would be the velocity?

 

[gneill]

so what would be the velocity?

Sorry, it is against the forum rules to simply provide answers to homework questions. You need to show your own work, which we can then help you with.

I would suggest that you start your own thread, filling out the homework help template, rather than resurrecting a nearly year-old thread; The Original Poster of this thread hasn't been logged in since March of 2017, so you're unlikely to receive a reply from him.

 

[conscience]

@gneill , Sorry if I am being a nuisance .I just wish to clarify a conceptual point in this question .

I think the book answer given in the OP is not correct .It is not the work done by the man . Rather it is the total work done on the (cart+man) system .

As far as work done by the man is concerned as asked in the question , it is same as the work done on the cart by the man . The result would be different from the one given in OP .It will be three times less than the book answer .

@TSny , @gneill , please correct me .

 

[gneill]

@gneill , Sorry if I am being a nuisance .I just wish to clarify a conceptual point in this question .

I think the book answer given in the OP is not correct .It is not the work done by the man . Rather it is the total work done on the (cart+man) system .

By conservation of momentum, the center of mass of the cart+man system does not move as there are no external forces operating, hence the net work done on the system is zero.

As far as work done by the man is concerned as asked in the question , it is same as the work done on the cart by the man . The result would be different from the one given in OP .It will be three times less than the book answer .

Within the system, work was done by the man on both himself and the cart since both obtained kinetic energy as a result of his actions.

 

[conscience]

By conservation of momentum, the center of mass of the cart+man system does not move as there are no external forces operating, hence the net work done on the system is zero.

In a non-rigid body or in a system comprising of different particles/bodies , net work done on the system and work done on center of mass are different things .

As you have also stated that both men and cart gain kinetic energies , means positive work has been done on both of them separately . Net work done is the sum total of work done on cart and man .

Internal forces do not change the motion of the center of mass but certainly work can be done by them .

Consider two masses moving towards each other under the influence of gravitational attraction .Positive work is done on each mass even though work done on the center of mass is zero .

Within the system, work was done by the man on both himself

I don't think a body can do work on himself . Work is generally defined for an external force .

But maybe you are right

I would like to read @TSny's opinion .

 

[TSny]

Interesting questions. Here's my take.

I agree that the work done by the man on the cart is 1/3 the answer given in the OP.

I tend to agree that the interpretation of "the total work done on the (cart+man) system" would be the total work done by external forces on the system. And this would be zero.

As you have also stated that both men and cart gain kinetic energies , means positive work has been done on both of them separately . Net work done is the sum total of work done on cart and man .

Actually, I think you can argue that negative work was done on the man! Before explaining that, it might be helpful to recall the example of a man standing on a frictionless floor and pushing on a wall. The man ends up sliding away from the wall with some kinetic energy. But no work was done on the man by the normal force from the wall, and no work was done on the wall by the force of the man on the wall. The gain in KE of the man comes from conversion of internal energy stored in the man's muscles, etc.

Going back to the man jumping off of the cart, the work done by the man's feet against the cart is positive and equals the gain in KE of the cart. The work done by the force of the cart on the man is negative. This is because the point of contact between the feet and the cart moves in the direction of the cart's motion while the force on the man from the cart is in the opposite direction. The work done by the cart on the man is the negative of the work done by the man on the cart.

Even though the work done on the man by the cart's force is negative, the man nevertheless gains KE. The KE of the man plus the KE of the cart equals the total internal energy consumed inside the man.

As we know, the "work-energy principle" (which states that the work done on an object equals the change in KE of the object) is only applicable to non-deformable bodies. So, it is not applicable to the man in this problem.

In summary, the work done by the man on the cart is 1/3 the answer in the OP. The answer given in the OP represents the total final KE of the system. This total KE does not come from any work done on the system (by external forces), but by conversion of the man's internal energy into KE of the system. It is hard to see how the answer in the OP could represent "the work done by the man".