A 0.34 kg mass sliding on a horizontal frictionless surface is attached to one e
ID: 2200627 • Letter: A
Question
A 0.34 kg mass sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 475 N/m) whose other end is fixed. The mass has a kinetic energy of 10 J as it passes through its equilibrium position (the point at which the spring force is zero). (a) At what rate is the spring doing work on the mass as the mass passes through its equilibrium position? 0 W (b) At what rate is the spring doing work on the mass when the spring is compressed 0.10 m and the mass is moving away from the equilibrium position? _________W I just need (b) thanks!Explanation / Answer
If x is the extension of the spring (x < 0 if it's compressed) then the conservation of the total mechanical energy can be stated this way, with v = velocity = dx/dt : E = 10 J = (1/2) [ m v^2 + k x^2 ] The spring force does work at the following rate: P = F (dx/dt) = -k x v = -k x sign(v) sqrt ( [2E-kx^2] / m ) "P" stands for "power" which is the name for the rate at which work is done. P is expressed in watts (W) or joules per second (J/s) same thing. (a) P is zero when x = 0 (Because the spring force is then zero and a zero force does no work.) (b) The situation described is x = -0.1 m and v < 0 (i.e., sign(v) = -1) P = -k x sign(v) sqrt ( [2E-kx^2] / m ) P = - (475 N/m)(-0.1 m) (-1) sqrt ( [20 - 475(0.1)^2] J / (0.34 kg) ) P = - 116.63... J/s = -116.63 W The negative sign says that the spring is working to slow down the mass because the force is opposite to the velocity. The answer is expressed with as many significant digits as the question.Related Questions
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