Work & Tension Homework: Find W & Delta K.E.

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SUMMARY

The discussion focuses on calculating the work done by a rope and the change in kinetic energy (delta K.E.) for a block with a mass of 4 kg, subjected to a tension of 35 N while being lowered 10 m. The work done by the rope is calculated using the formula W = Fd, resulting in -350 J, indicating that the rope does negative work on the block. The participants clarify the relationship between gravitational potential energy and kinetic energy, emphasizing the importance of sign conventions in these calculations.

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Homework Statement



There is a block with a mass of 4kg attached to a rope where tension is point upwards. The block is lowered by the rope, and the tension is 35N. If block is lowered 10m:

A) What is work done by the rope?
B) What is delta K.E.?

Homework Equations



W = mgh
W = Fd

The Attempt at a Solution



A) W = Fd = -350J

I am having a hard time with the concept as to why the block does not affect the work on the rope. Also, how is the block losing energy (indicated by the negative sign in the solution)?

B) Can I solve it this way:

Fd = delta U + delta K

(35)(10) = (4)(10)(10) + delta K

-50J = delta K
 
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Gear2d said:

Homework Statement



There is a block with a mass of 4kg attached to a rope where tension is point upwards. The block is lowered by the rope, and the tension is 35N. If block is lowered 10m:

A) What is work done by the rope?
B) What is delta K.E.?

Homework Equations



W = mgh
W = Fd
You're missing a trig factor here (which gives you the minus sign in part a).

The Attempt at a Solution



A) W = Fd = -350J

I am having a hard time with the concept as to why the block does not affect the work on the rope. Also, how is the block losing energy (indicated by the negative sign in the solution)?
If you compare a block being lowered by this rope to a block that is just dropped 10m, the one that is dropped has more energy after dropping 10m.

B) Can I solve it this way:

Fd = delta U + delta K

(35)(10) = (4)(10)(10) + delta K

You have some sign errors here. You had already found in part a that the work was -350J. For the delta U term, what is delta h?
 

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