Calculating Speed on an Inclined Plane | Mechanical Energy #2

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To calculate the speed of a 19 kg box sliding down a frictionless ramp, apply the conservation of mechanical energy principle. The potential energy (PE) at the top, calculated as PE = mgh, equals the kinetic energy (KE) at the bottom, expressed as KE = 1/2 mv^2. Setting these two equations equal allows for solving the speed at the bottom of the ramp. The final velocity can be determined using the formula v = √(2gh), where g is the acceleration due to gravity. Ensure the answer is provided with the correct units.
billyghost
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A 19 kg box starts at rest and slides down a frictionless ramp. The length of the ramp is 4.5 m and the height above the ground at the top is 1.3 m. How fast is the box moving at the bottom of the ramp?

Unsure of inclined planes...help with formulas, etc.?
 
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Think conservation of mechanical energy (KE + PE). (Measure the potential energy from the bottom of the ramp.)

Mechanical Energy (at top of ramp) = Mechanical Energy (at bottom of ramp)
 
19 x 1.3 x 9.8 ?
 
billyghost said:
19 x 1.3 x 9.8 ?
That looks like a calculation of the PE at the top of the ramp: PE = mgh. (The units will be Joules.) Which happens to be the total mechanical energy, since it starts from rest. Since energy is conserved, this also equals the KE at the bottom of the ramp. ({KE} = 1/2 m v^2)

So set the PE at the top (mgh) equal to the KE at the bottom (1/2 m v^2) and solve for the speed.
 
Therefore,velocity should be the square root of 2 x 9.8 x 1.3
 
Right. But be sure to give your answer with the proper units.
 
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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