Mechanical Energy Homework: Find E of 0.13 kg Ball at 5.00 m

AI Thread Summary
To find the mechanical energy of a 0.13 kg ball at 5.00 m above the ground and moving at 20.0 m/s, both kinetic and potential energy must be considered. The kinetic energy (Ek) is calculated using the formula Ek = 1/2mv^2, resulting in 26 J. The potential energy (Ep) is determined using Ep = mass x gravity x height, yielding 6.37 J. The total mechanical energy is the sum of kinetic and potential energy, which equals 32.4 J. Understanding the inclusion of potential energy is crucial for solving similar problems.
cybernerd
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Homework Statement



A 0.13 kg ball moves at a rate of 20.0 m/s at the point where it is 5.00 m above the ground. How much mechanical energy does the ball have with respect to the ground?

Homework Equations



Ek = 1/2mv^2

The Attempt at a Solution



Ek = 1/2mv^2
Ek = 1/2(0.13kg)(20m/s)^2
E = 26 J

The assignment says my answer should be 32.4 J.
Obviously, I am doing something wrong, but I'm not sure how to incorporate the "with respect to the ground" bit into my equations.

Somebody offer me some guidance, please?
 
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I can't comment on WHY this is so.. but apparently.. the answer includes the potential energy of the ball as well

= mass x gravity x height

= .13 * 9.80 * 5m = 6.37J

26 J + 6.37J = 32.4J



Not really helping but I'm not familiar with mechanical energy...
 
Hmm...that is strange...I really don't like the way that the question is worded.

But that did help, thank you.
 
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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