A 0.35 kg mass sliding on a horizontal frictionless surface is attached to one e

Question

A 0.35 kg mass sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 750 N/m) whose other end is fixed. The mass has a kinetic energy of 9.0 J as it passes through its equilibrium position (the point at which the spring force is zero). At what rate is the spring doing work on the mass as the mass passes through its equilibrium position?

At what rate is the spring doing work on the mass when the spring is compressed 0.077 m and the mass is moving away from the equilibrium position?

Explanation / Answer

At what rate is the spring doing work on the mass when the spring is compressed 0.077 m and the mass is moving away from the equilibrium position?

My Answer:
Let:
v be the velocity,
x be the displacement,
w be the angular frequency,
a be the amplitude,
k be the spring constant,
K be the kinetic energy at displacement x,
K0 be the kinetic energy at the equilibrium position,
P be the required power.

w^2 = k / m
v^2 = w^2(a^2 - x^2)
= k(a^2 - x^2) / m
v = sqrt[ k(a^2 - x^2) / m ] ...(1)

K = mv^2 / 2
= k(a^2 - x^2) / 2 ...(2)

Putting x = 0 and K = K0:
K0 = k a^2 / 2 ...(3)
a^2 = 2K0 / k ...(4)

Eliminating a^2 from (2) and (3):
K = k(2K0 / k - x^2) / 2
= K0 - kx^2 / 2

P = dK / dt = - kx dx / dt
= - kxv

Substituting for v from (1):
P = - kx sqrt[ k(a^2 - x^2) / m ]

Substituting for a^2 from (4):
P = - kx sqrt[ k(2K0 / k - x^2) / m ]
= - kx sqrt[ (2K0 - kx^2) / m ]
= - 750 * 0.077 sqrt[ (2 * 9.0 - 750 * 0.077^2) / 0.35 ]
= - 2236 watt

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