<p>A shopper in a supermarket pushes a cart with a force of 35 N directed at an
ID: 2005125 • Letter: #
Question
<p>A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25° below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at constant speed. (a) Find the work done by the shopper as she moves down a 50.0 m length aisle. (b) What is the net work done on the cart? (c) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doen't change, would the shopper's applied force be larger, smaller, or the same? What about the work done on the cart by the shopper?</p>Explanation / Answer
The work done on the cart is equal to the various forces times the distance through which the cart moves. Conceptually, it helps to realize that since the energy of the cart is not changing as its velocity is not changing, the net work done by the various forces must sum to zero because doing net work on a system would change its energy. As the only forces acting on the cart are the pushing and friction, these must cancel out, i.e. be of equal magnitude but in opposite directions. To find the work done by the pushing force, first find the component of the force parallel to the direction of motion, or the x component of the force. To do this set up a right triangle with the force on the hypoteneuse with the hypoteneuse 25 degrees above the x-component. Doing a little trig gives you the result that the x-component of the force equals the total force times the cosine of the angle, or Fx=F*cos(theta) Now that you have the force supplied by the shopper in the x direction, it is easy to find the work done by multiplying the force times the displacement in the x direction, W=Fd You have the force and you have the distance, now simply multiply and find work. For part b, as I already mentioned, the net force on the cart is zero, and moreover there is no change in the energy of the cart and so the net work must be zero as well. Keep in mind this is the NET work and NET force of which i am talking, the pushing is doing positive work as it is in the same direction as displacement, and friction which is causing a force in the opposite direction is doing work of equal and opposite magnitude which nets to zero. In part c, the only thing that changes is the angle the shopper is making with the cart, this means that more of the force is directed in the x-axis and less in the y-axis which means that not only does the force in the x-direction increase, but the decreased part in the y-axis decreased friction. If the shopper maintained the same total force, the cart would accelerate as the force supplied would now be great enough to overcome friction, so to maintain the same speed, he must supply less force.
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