Question

A 1.40 kg block is attached to a spring with spring constant 15.5 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 43.0 cm/s . What are The block's speed at the point where x= 0.650 A

Answer 1

Answer:

v = 3.08 m/s.

Explanation:

Given,

• Initial velocity of the block = u = 43 cm/s = 0.43 m/s
• mass of the block = m = 1.40 kg
• spring constant = k = 15.5 N/m.
• x = 0.650 m

At x = 0.650 m

Let 'a' be the acceleration of the block at x.

Total force due to the spring force on the blcok at x= kx

$\therefore F_s\ =\ kx\\\Rightarrow ma\ =\ kx\\\Rightarrow a\ =\ \dfrac{kx}{m}\\\Rightarrow a\ =\ \dfrac{15.5\times 0.650}{1.40}\\\Rightarrow a\ =\ 7.19\ m/s^2$

Hence the acceleration of the block at x is 7.19\ m/s^2.

Let 'v' be the velocity of the block at x.

From the kinematics,

$v^2\ =\ u^2\ +\ 2ax\\\Rightarrow v\ =\ \sqrt{u^2\ +\ 2ax}\\\Rightarrow v\ =\ \sqrt{0.43^2\ +\ 2\times 7.19\times 0.650}\\\Rightarrow v\ =\ 3.08\ m/s.$

Hence at the compression x = 0.650 m the velocity of the block is 3.08 m/s.