Question

The amplitude of the subsequent oscillauohs Express your answer with the appropriate units. Review A 1.40 kg block is attached to a spring with spring constant 15.0 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 40.0 cm/s What are 12.2 cm Previous Answers You may want to review (Pages 400-401) Correct Here we learn how to calculate the amplitude of the spring's oscillations Part B The block's speed at the point where 0.550 A? Express your answer with the appropriate units. cm 400 Ls SubmitPre Previous Answers Request Answer X Incorrect; Try Again; 19 attempts remaining

Answer 1

In this given question, m = 1.40 Kg and spring constant, K = 15.0 N /m.

Now, from the law of conservation of energy, we have:

Kinetic energy gained by block = Potential energy stored in the spring.

Now, Kinetic energy gained by block = 1/2*m*v^2 where m = mass and v = Speed

Given: V = 40 cm/s or 0.4 m/s.

Putting the values- K.E. = 1/2 * 1.40 * 0.4^2

And we know, Potential Energy = 1/2Kx^2

=> P.E = 1/2* 15 * x^2

but K.E = P.E ( as mentioned above)- K.E- Kinetic Energy and P.E is Potenital Energy

Thus solving it to get the value of x

=> 1/2 * 1.40 * 0.4^2 = 1/2 *15 * x^2

=> x^2 = 0.0149 or x = 0.1222m. or 12.2 cm- Part A

Part B - Block speed at the point when x = 0.550A or 0.550*0.122

Now, Initial Kinetic energy = Kinetic energy at x + Potential energy at x

=> 1/2mV^2 = 1/2mv^2 + 1/2Kx^2

=> 1/2 *1.40*0.4^2 = 1/2 * 1.40 * v^2 + 1/2 * 15.0 * (0.550 * 0.122)^2

=> v = 0.3343 m/s. -Answer