Question

A block of mass m is compressed against a spring (spring constant kk) on a horizontal frictionless surface. The block is then released and subsequently falls from a high ground of height hh. The maximum compression of the spring before the block was released is \Delta xΔx. The velocity of the block after release is v_1v​1​​. The velocity of the block when it hits the ground is v_2v​2​​.

A.If the block started with a compression of \frac{1}{2}\Delta x​2​​1​​Δx, the block velocity after leaving would be \frac{1}{2}v​2​​1​​v.
B.The mechanical energy of this system is not conserved.
C.If height h is changed to h=2hh=2h, then the velocity of the block v_2v​2​​ will be changed into v_2=2vv​2​​=2v.
D.If the block started with a compression of 3\Delta x3Δx, the block’s velocity after leaving would be 6v6v.
E.None of the above statements are true.

Answer 1

A) True, B) False, C) False and D) false

Step-by-step explanation:

Let's solve the problem using the law of conservation of energy to know if the statements are true or false

Let's look for mechanical energy

Initial

Emo = Ke = ½ k Dx2

Final

Em1= ½ m v12

Emo = Em1

½ k Δx2 = ½ m v₁²

v₁² = k / m Δx²

v₁ = √ k/m Δx

Now let's calculate the speed when it falls

Vfy² = Voy² - 2gy

Vfy² = - 2gy

Vf² = v₁² + vfy²

A) True v₁ = A Δx

.B) False. As there is no rubbing the mechanical energy conserves

.C) False the velocity is proportional to the square root of the height

v2y = v2 √2

. D) false promotional compression speed