0.25 kg mass sliding on a horizontal frictionless surface is attached to one end

Question

0.25 kg mass sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 850 N/m) whose other end is fixed. The mass has a kinetic energy of 11.0 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 0.080 m and the mass is moving away from the equilibrium position?

Explanation / Answer

a)

At equilibrium Position :

x = 0

hence , spring force = Fs = kx = 850 (0) = 0 N

V = speed at equilibrium position

rate of work = Fs V = 0            since Fs = 0 N

b)

x = 0.08 m

Fs = k x = 850 x 0.08 = 68 N

spring potential energy is given as

SPE = (0.5) k x2 = (0.5) (850) (0.08)2 = 2.72 J

Using conservation of energy

SPE at X = 0.08   +   KE at X = 0.08    = KE at X = 0

2.72 + (0.5) m V'2 = 11

2.72 + (0.5) (0.25) V'2 = 11

V' = 8.14 m/s

Rate of doing work = -Fs V' = - 68 x 8.14 = 68 x 8.14 = - 553.52

Related Questions

Navigate

• Browse (All)
• Browse
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.