A 0.52-kg object connected to a light spring with a force constant of 19.4 N/m o

Question

A 0.52-kg object connected to a light spring with a force constant of 19.4 N/m oscillates on a frictionless horizontal surface. The spring is compressed 4.0 cm and released from rest. (a) Determine the maximum speed of the object. Correct: Your answer is correct. m/s (b) Determine the speed of the object when the spring is compressed 1.5 cm. Correct: Your answer is correct. m/s (c) Determine the speed of the object as it passes the point 1.5 cm from the equilibrium position. Correct: Your answer is correct. m/s (d) For what value of x does the speed equal one-half the maximum speed? Incorrect: Your answer is incorrect.

Explanation / Answer

Using Energy Conservation,

Potential Energy + Kinetic Energy = E (constant)

At the beginning

Kinetic Energy = 0 and Potential Energy = 0.5kx2, where k = 19.4 N/m and x = 0.04 m

So,

0.5*19.4*(0.04)2 + 0 = E

--> E = 0.01552 J

a) Max Speed when potential energy is 0 (i.e. x = 0). This gives

0 + KEmax = E
KEmax = 0.01552 J
0.5mv2 = 0.01552 J

0.5*0.52* v2 = 0.01552

vmax = 0.244 m/s

b) x = 0.015

E = 0.5mv2 + 0.5kx2

0.01552 = 0.5*0.52*v2 + 0.5*19.4*(0.015)2

This gives v = 0.226 m/s = 0.23 m/s

c) same as b

d) 0.5m(vmax/2)2 + 0.5kx2 = 0.01552 J

0.5*0.52*(0.244/2)2 +0.5*19.4*x2 = 0.01552

Solving, we get x = 0.0346 m = 3.5 cm

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