A 1.20 block is attached to a spring with spring constant 14.0 . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 50.0 . What is The block's speed at the point where = 0.350 A?
To determine the block's speed at the point where the displacement (x) is 0.350A, we can use the principle of conservation of mechanical energy. The mechanical energy of the system (block and spring) remains constant throughout the motion.
The mechanical energy of a block-spring system is the sum of kinetic energy (KE) and potential energy (PE) due to the spring. Mathematically, it can be represented as:
E = KE + PE
Initially, when the block is at rest, it has no kinetic energy, and the entire energy is stored as potential energy in the spring. Therefore, initially, E = PE.
At any point during the motion, the potential energy of the system can be calculated using Hooke's Law:
PE = (1/2)kx^2
where k is the spring constant and x is the displacement from the equilibrium position.
The kinetic energy of the block can be given by:
KE = (1/2)mv^2
where m is the mass of the block and v is the velocity of the block.
Now, at the point where x = 0.350A, we need to calculate the final velocity v.
Using the conservation of mechanical energy, we can equate the initial potential energy to the final sum of kinetic and potential energy:
PE_initial = KE_final + PE_final
Initially, the potential energy is stored in the spring:
PE_initial = (1/2)kA^2
At the point where x = 0.350A, the potential energy will be:
PE_final = (1/2)k(0.350A)^2
The final kinetic energy can be calculated by rearranging the equation:
KE_final = PE_initial - PE_final
The final velocity can be calculated using the equation for kinetic energy:
v = sqrt((2/m) * KE_final)
Substituting the values given in the problem:
m = 1.20 kg (mass of the block)
k = 14.0 N/m (spring constant)
A = amplitude
You need to substitute the value of the amplitude (A) given in the problem to calculate the final velocity (v) at x = 0.350A.