A 0.200-kg object is attached to a horizontal, light spring that has a force constant of 75.0 N/m...

Question

A 0.200-kg object is attached to a horizontal, light spring that has a force constant of 75.0 N/m. The object is pulled 8.00 cm to the right of equilibrium and released from rest to slide on a horizontal table that exerts negligible friction on the object.

a. Calculate the maximum kinetic energy of the object.

1. I understand that I use 1/2kx^2=1/2mv^2 to get max Ke = 0.24J

b. Find the location of the object when the speed is equal to one-third of the maximum speed, the object is moving to the right, and the acceleration is positive.

c. Calculate the value of the spring potential energy at this point.

Answer 1

a) You are correct. The maximum potential energy occurs when the spring is stretched right by 8 cm. But, since we can easily get goofed up by units here, let's keep everything in SI units and call it 0.08 m. When released, the mass will reach maximum velocity when it crosses the equilibrium position

(1/2)kx² = (1/2)mv²

kx² = mv²

v² = (kx²/m)

v_max = √(kx²/m)

v_max = x√(k/m) = 0.08√(75/0,2) = 0.08√375 = 1.549 ≈ 1.55 m/s

b) The equation for position for the block is given generally in the form:

x(t) = Acos(ωt) where A is the amplitude which is the amount you stretch it (so in this case A = 0.08) and ω is the angular frequency which equals √(k/m) (so in this case ω = √(75/0.2) = √375 ≈ 19.365

That makes the equation of motion:

x(t) = 0.08cos(19.365t)

The equation for the velocity for the block is given by v(t) = -Aωsin(ωt) so, plugging in the values for this problem: v(t) = -0.08(19.365)sin(19.365t) or:

v(t) = -1.549sin(19.365t)

Since Vmax = 1.549 then a third of that is 0.5164 m/s. Thinking about this velocity in terms of the statement "the object is moving to the right, and the acceleration is positive", this would mean the abject has left the starting line, traveled left past the equilibrium position, past the far left position and is now approaching the equilibrium position again (moving right) meaning velocity is positive. To find the time, plug in v = 0.5164:

0.5164 = -1.549sin(19.365t)

-1/3 = sin(19.365t)

sin^-1(-1/3) = sin^-1(sin(19.365t))

Plugging in sin^-1(-1/3) into the calculator, we get -0.33984 BUT this is a little problematic in this case because it gave us the result in Q4 (typical answer for sin^-1) BUT, we are looking for the answer in Q3 to correspond to the description we got so we need to take the value +0.33984 and add it to π to get he equivalent position in Q3 giving us 3.4814. To verify, try sin(3.4814) and you see you get -1/3. So:

3.4814 = 19.365t

t = 0.17978 s

Now to find the location, plug that value of "t" into the position equation:

x(0.17978) = 0.08cos(19.365•0.17978) = -0.0754 cm or 7.54 cm left of the equilibrium position

To get the value of the spring potential energy at this point, you can use E_P = 1/2kx²

E_P = (1/2)(75)(0.0754)² = 0.2133 j

We can check that answer by taking the total energy (1/2)(75)(0.08)² = 0.24 j and subtract off the kinetic energy (1/2)mv² = (1/2)(0.2)(0.5164)² = 0.0267:

0.24 - 0.0267 = 0.2133 j so it checks out.