Question

After hitting the spring, the block is bounced back up the ramp. The maximum compression of the spring is Δx=0.03m, and the spring constant is k=800N/m. What is the velocity of the block when it first reaches a height of 0.1m above the ground on the ramp? The mass of the block is m=0.2kg, take g=9.8m/s^2 .

Answer 1

Answer:

v = 1.28 m/s

Explanation:

Given that,

Maximum compression of the spring, $\Delta x=0.03\ m$

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

$\dfrac{1}{2}kx^2=mgh+\dfrac{1}{2}mv^2$

Substituting all the values in above equation,

$\dfrac{1}{2}\times 800\times 0.03^2=0.2\times 9.8\times 0.1+\dfrac{1}{2}\times 0.2\times v^2$

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.