Answer:
v = 1.28 m/s
Explanation:
Given that,
Maximum compression of the spring, [tex]\Delta x=0.03\ m[/tex]
Spring constant, k = 800 N/m
Mass of the block, m = 0.2 kg
To find,
The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.
Solution,
When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,
Initial energy = Final energy
[tex]\dfrac{1}{2}kx^2=mgh+\dfrac{1}{2}mv^2[/tex]
Substituting all the values in above equation,
[tex]\dfrac{1}{2}\times 800\times 0.03^2=0.2\times 9.8\times 0.1+\dfrac{1}{2}\times 0.2\times v^2[/tex]
v = 1.28 m/s
Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.
