What is the work done on cart by the string?

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SUMMARY

The work done on the cart by the string in a system involving a cart of mass M1 = 6 kg and a block of mass M2 = 3 kg can be calculated using the equation Ws = M2gh, where h = 1 m. The system is frictionless, and both the cart and block share the same acceleration and velocity due to the tension in the string being equal. The total work done on the cart is directly related to the gravitational potential energy lost by the block as it falls.

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Homework Statement


A cart of mass M1 = 6 kg is attached to a block of mass M2 = 3 kg by a string that passes over a frictionless pulley. The system is initially at rest and the table is frictionless. After the block has fallen a distance h = 1 m:

What is the work Ws done on the cart by the string?
Ws =

Homework Equations


Work done by F = F(cosθ)d
Work done total = ΔKE object
KE = 1/2mv^2
Wsystem = Wg(M2) = Fgh = M2gh

The Attempt at a Solution


I know that the velocity and acceleration of cart must be the same as velocity and acceleration of block. I also know that the tension of the cart and block should be equal. But how do you come up with the final expression to find the work done on the cart by the string.
 
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Hello again, hope you benefited from earlier assistance ... :rolleyes:

In this case you should post your work -- I suspect you already have worked out quite a bit of what's needed for this exercise :smile:
 

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