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 November 12th, 2010, 12:34 AM #1 Senior Member   Joined: Oct 2009 Posts: 883 Thanks: 1 A shopper in a supermarket pushes a cart with a force [color=#0040FF]1 ) A shpooer in a supermarket pushes a cart with a force of 25 N directed at an angle of 18 downward from the horizontal . Find the work done by the shopper as she moves a 40 m lenght of the aisle ?[/color] W = (35 N)(40) cos 25o W = (35 N)(40.906) W = 1 586 N-m = W = F d cos 25 = 1 586 J [color=#0040FF]2 ) Tarzan swing on a 25 m long vine initially inclined at angle of 32 with the vertical . What is the speed at the bottom of the swings a) if he start from rest ? b) if he pusches off with speed of 5.00 m/s ?[/color] a ) if he start from rest ? = zero
 November 12th, 2010, 06:03 AM #2 Global Moderator     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 11,926 Thanks: 316 Math Focus: The calculus Re: A shopper in a supermarket pushes a cart with a force 1.) First, we need to find the horizontal component of the shopper's pushing force. This is given by: $F_x=F\cdot\cos(\theta)$ where $\theta$ is the angle of inclination. For a constant force along a line, we know: The work W done by an agent exerting a constant force is the product of the component of the force in the direction of the displacement and the magnitude of the displacement of the force. Thus, in this situation, we have: $W=F_x\cdot x$ $F_x=((25\text{ N.})(\cos(-18^{\circ})))(40\text{ m.})=250\sqrt{2(\sqrt{5}+5)}\text{ J.}\approx 951.1\text{ J.}$ 2. By applying the principle of constancy of mechanical energy, we can derive a formula to answer both parts of the question. The sum of the initial kinetic and potential energies will equal the sum of the final kinetic and potential energies. If we measure the y-coordinates from the center of rotation (which we take as the origin), then: $y_i=-L\cos(\theta_i)$ $y_f=-L$ Therefore: $U_i=-mgLcos(\theta_i)$ $U_f=-mgL$ $K_i=\frac{1}{2}mv_i^2$ $K_f=\frac{1}{2}mv_f^2$ Using the constancy of energy, we have $K_i+U_i=K_f+U_f$ $\frac{1}{2}mv_i^2-mgLcos(\theta_i)=\frac{1}{2}mv_f^2-mgL$ Multiply through by $\frac{2}{m}$: $v_i^2-2gLcos(\theta_i)=v_f^2-2gL$ Solve for $v_f$: $v_f=\sqrt{v_i^2-2gLcos(\theta_i)+2gL}$ $v_f=\sqrt{v_i^2+2gL\left(1-cos(\theta_i)\right)}$ Now, all you need to do is plug in the known and given data: $g=9.81\text{ \frac{m.}{s.^2}}$ $L=25\text{ m.}$ $\theta_i=32^{\circ}$ for (a) $v_i=0\text{ \frac{m.}{s.}}$ and for (b) $v_i=5.00\text{ \frac{m.}{s.}}$

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