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Monday, October 5, 1998 Chapter 5: Springs Chapter 6: Linear Momentum Conservation of Momentum Impulse Let’s first figure out the force delivered by the motor... A 600 kg elevator starts from rest and is pulled upward by a motor with a constant acceleration of 2 m/s2 for 3 seconds. What is the average power output of the motor during this time period? Fnet = ma = (600 kg)(2 m/s2) Fnet = 1200 N = Fmotor - Fg = Fmotor - mg 1200 N = Fmotor - (600 kg)(10 m/s2) = Fmotor - 6000 N Fmotor = 6000 N + 1200 N = 7200 N Now we need to determine the work done by the motor... W = F Ds A 600 kg elevator starts from rest and is pulled upward by a motor with a constant acceleration of 2 m/s2 for 3 seconds. What is the average power output of the motor during this time period? But we don’t know Ds, so…. s = s0 +v0t +0.5at2 = 0 + 0 + 0.5(2 m/s2)(3 s)2 = 9 m W = (7200 N)(9 m) = 64800 J W 64800 J P 21600 W Dt 3s In addition to the gravity, there are other mechanisms to store POTENTIAL ENERGY. One of them is... Sir Robert Hooke unlocked the secret of the spring... A spring resting in its natural state, with a length l exerts no horizontal force on anything! However, if we compress or stretch the spring by some amount x, then the spring is observed to exert a Force in the opposite direction. x l Hook discovered this force could be modeled by the mathematical expression F = - kx Notice that this force operates along a linear line! Force Which means that if we looked at the plot of Force versus compression/stretching x... x Slope of this line is -k, where k is the spring constant. Force If we look at the work done by an applied force which compresses the spring through a distance (-x1)... F1 W FDs Fx1 ( F1 0) x 1 2 W ( k )( x1 ) x1 kx 1 2 1 2 2 x -x1 Work done BY the external force ON the spring. This energy is stored in the spring... Potential Energy of a spring is PE spring 1 2 kx 2 So, for spring problems, we have a new TOTAL MECHANICAL ENERGY given by KE PE g PEspring And it is THIS quantity which will be conserved absent other, outside forces. Momentum & Collisions The linear momentum of an object of mass m moving with velocity v is defined as the product of the mass and the velocity: p mv p mv Notice that momentum is a vector quantity, which means that it must be specified with both a magnitude and direction. Also notice that the direction of the momentum vector is necessarily parallel to the velocity vector. p mv [ p] [m][v ] [ p] kg (m / s) OR [ p] N s The units suggest a relationship between force and momentum. p mv What happens when we apply a force to an object? It accelerates. Its velocity changes. Its momentum changes. The force imparts momentum. p mv By how much will the momentum change? That depends upon the length of time over which the force is applied to the object. v v0 aDt mv mv0 ma Dt mv mv0 FDt mv mv0 FDt p p0 FDt Change in momentum Dp FDt I Impulse The impulse of a force on an object equals the change in momentum of that object. Notice that impulse is a vector quantity as well!