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

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The discussion focuses on calculating the work done by a rope and the change in kinetic energy (delta K.E.) for a block being lowered. The work done by the rope is calculated as W = Fd, resulting in -350J, indicating that the rope does negative work on the block as it loses energy. The confusion arises regarding the block's energy loss and the negative sign, with comparisons made to a block dropped freely. For delta K.E., the approach involves using the relationship between work, potential energy, and kinetic energy, but sign errors in calculations are noted. Understanding these concepts is crucial for accurately solving the problem.
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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?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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