Energy Conservation & Potential Energy Spring/Friction Problem

AI Thread Summary
A 0.80-kg block is projected from a spring after an external force is removed, achieving initial and final velocities of 1.2 m/s and 1.9 m/s, respectively, on a frictionless ramp. Upon entering a rough section, the block's velocity decreases to 1.4 m/s due to kinetic friction with a coefficient of 0.39. The discussion revolves around applying energy conservation principles to solve for the spring's force constant, k, despite the absence of distance measurements. Participants suggest using the equations for kinetic and potential energy to derive k, but express confusion regarding how to proceed without distance values. The conversation emphasizes the relationship between energy equations and the need for additional information to solve the problem effectively.
Help1212
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Homework Statement



A 0.80-kg block is held in place against the spring by a 67-N horizontal external force. The external force is removed, and the block is projected with a velocity v1 = 1.2 m/s upon separation from the spring. The block descends a ramp and has a velocity v2 = 1.9 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction is 0.39. The velocity of the block is v3 = 1.4 m/s at C. The block moves on to D, where it stops. In Fig. 7.4, the force constant of the spring is..

7.4.jpg


Homework Equations



K1 +U1 +W other = K2+ U2.. ect .. U2=0.5kx^2 .. K1=(0.5)mv^2

The Attempt at a Solution



I've tries solving the problem with the equation above but I really just don't know where to start especially since there are no units of distance given.. Can someone lead me in the right direction please?
 
Last edited:
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Hi Help1212! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)

I assume you can find kx and kx2/2 ?

ok, then k = … ? :wink:
 
euhh... I'm still a little confused..how do i find k if I don't know x?
 
x is twice kx2/2 divided by kx :wink:
 
hmm. well when I move around the equation i get k=2[0.5(mv22)/x2

from the assumption that K1 + U1 = K2 + U2

Am i going in the right direction??
 
can you find kx and kx2/2 ?

if so, what are they?
 
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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