A 0.30 kg block sliding on a horizontal frictionless surface with a speed of 2.5
ID: 2222741 • Letter: A
Question
A 0.30 kg block sliding on a horizontal frictionless surface with a speed of 2.5 m/s strikes a light string that has a spring constant of 3.0 x10^3 N/m. What is the total mechanical energy of the system? What is the kinetic energy K of the block when the spring is compressed a distance x=1.0 cm? (assume that no energy is lost in the collision) How far will the spring be compressed when the block comes to a stop? (solve using energy principles.) Please explain how to do this if possible. I am trying to understand how to do these problems and it doesn't really help just seeing the formulas and numbers plugged in. I want to grasp it conceptually. Thank you.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 = - (500 N/m)(-0.1 m) (-1) sqrt ( [20 - 500(0.1)^2] J / (0.30 kg) ) P = - 353.55339... J/s = -350 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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