How High Will the Block Rise After Hitting the Spring?

AI Thread Summary
The discussion focuses on calculating how high a 2.0 kg block will rise after hitting a spring, given its initial speed of 11 m/s and a spring constant of 4000 N/m. Using the conservation of energy principle, the initial kinetic energy of the block is calculated to be 121 J. This energy is converted into potential energy in the spring, leading to a maximum displacement of approximately 0.077 m. By subtracting this displacement from the initial height of 18 meters, the block's final height after leaving the spring is determined to be about 17.92 meters. Thus, the block will rise to a height of approximately 17.92 meters after the interaction with the spring.
Hughey85
Messages
14
Reaction score
0
A 2.0 kg block is falling at a speed of 11 m/s and is 18 meters above the spring. The spring constant is 4000 N/m, to the nearest hundredth of a meter what height will the block rise after it hits and leaves the spring?


Help please! I can find the kinetic energy of the falling block by using... K=.5mv^2 ... I don't know how to go from there! Help!
 
Last edited:
Physics news on Phys.org
Law of conservation of energy :
The total energy at the beginning = The total energy at the end
\frac{1}{2}mv^2+mgh_1=mgh_2
where
v=the block initial speed(initial state)
h1=the block height relative to the spring(initial state)
h2=the maximum height the block will reach (relative to the spring)ie v =0 (final state)
 


To find the height the block will rise after hitting the spring, we can use the conservation of energy principle. This principle states that the total energy of a system remains constant, so the initial kinetic energy of the falling block will be equal to the potential energy of the block and the spring when it reaches its maximum height after being compressed by the spring.

First, we can calculate the initial kinetic energy of the block using the formula K = 0.5mv^2, where m is the mass of the block (2.0 kg) and v is the velocity (11 m/s). Plugging in these values, we get K = 121 J.

Next, we can calculate the potential energy of the block and spring system at the maximum height using the formula U = 0.5kx^2, where k is the spring constant (4000 N/m) and x is the maximum displacement of the spring. We can assume that all of the kinetic energy of the block is converted into potential energy of the block and spring system, so we can set K = U.

Substituting the values and solving for x, we get x = √(2K/k) = √(2(121 J)/(4000 N/m)) ≈ 0.077 m. This is the maximum displacement of the spring when the block reaches its maximum height.

Since the block was initially 18 meters above the spring, we can subtract the maximum displacement of the spring from the initial height to find the final height of the block. So, the block will rise to a height of 18 m - 0.077 m = 17.923 m after it hits and leaves the spring.

To summarize, the block will rise to a height of approximately 17.92 meters after hitting and leaving the spring.
 
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}...
Back
Top