Mechanical Energy & Speed of 48kg Block on 188m Hill

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A 48.0 kg block slides along the frictionless surface of a hill that is 188 m high with an initial speed 19.0 m/s. Relative to the bottom of the hill, (Note that the acceleration is not constant so you can not use freefall equations to solve the following problems.)

What is the block's mechanical energy when the block is 75.0 m above the bottom of the hill?

What is the block's speed when the block is 75.0 m above the bottom of the hill?
 
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you should be able to use conservation of energy seeing that the slope is frictionless, and all you need to know are the horizontal and vertical initial speeds.
 
i know the equations:

ME=KE+PE
KE= 1/2 mv^2
PE= mgh

my problem is that i don't know how to use them and i need someone to explain it to me.
 
You have total energy of block is: E = K + U. Base on conservation of energy, you have E0 = E1 <==> K0 + U0 = K1 + U1. You already have initial velocity ==> K0, height of hill = 188m ==> U0, height of hill at time t 75m ==> U1. Plug them to CE equation, you will find K1.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
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