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Work and net work

by mandy9008
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mandy9008
#1
Jun21-10, 05:53 PM
P: 127
1. The problem statement, all variables and given/known data
A shopper in a supermarket pushes a cart with a force of 39 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 48 m length aisle.
(b) What is the net work done on the cart? Why?
(c) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesn'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?


2. Relevant equations
W=Fxcosө


3. The attempt at a solution
a. W=39N (48m) cos-25
W=1696.6J

b. I thought that this would also be 1696.6J because work done by the shopper equals the net work done on the cart
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Doc Al
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Jun21-10, 06:06 PM
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Quote Quote by mandy9008 View Post
b. I thought that this would also be 1696.6J because work done by the shopper equals the net work done on the cart
Net work means the work done by all forces acting on the body. The force that the shopper exerts is only one of the forces acting. (What must the net force be on the cart?)
mandy9008
#3
Jun21-10, 06:09 PM
P: 127
So there are gravitational forces on the cart as well?

Doc Al
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Jun21-10, 06:27 PM
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Work and net work

Quote Quote by mandy9008 View Post
So there are gravitational forces on the cart as well?
Sure, but they aren't important. (Since gravity acts perpendicular to the direction of motion, it does no work.)

Hint: "the cart moves at constant speed" What does that tell you about the net force?
mandy9008
#5
Jun21-10, 08:54 PM
P: 127
if it is moving at a constant speed then the net work will be zero, right? I remember my professor saying that in class, but he never explained why. I am guessing that there is some force that is equal and opposite of the work done by the shopper, which could be explained by Newton's third law
collinsmark
#6
Jun21-10, 10:48 PM
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Hello mandy9008,

Quote Quote by mandy9008 View Post
if it is moving at a constant speed then the net work will be zero, right? I remember my professor saying that in class, but he never explained why.
Be careful how you interpret that, but yes, it is essentially true if you consider all the relevant individual forces involved, even frictional forces if they apply (and they do in this problem), and potential energy related forces (which don't apply in this problem).

When contemplating the issue, ask yourself these questions:

(a) If an object is moving at a constant velocity, what does that tell you about its acceleration a?

(b) Knowing that Fnet = ma, and if you also happen to know that a = 0, what does that tell you about Fnet (where Fnet can be the sum of multiple forces)? Thus what can you conclude about all the individual forces that make up Fnet? (Hint: if you were thinking that it means each of the individual forces making up Fnet is zero, that's not generally true. But you can say something about how they combine together.)

(c) If an object does not change velocity, what does that tell you about the object's change in kinetic energy?

(d) If an object's kinetic energy and potential energy do not change, what does that say about the total mechanical work done on that object by the all forces acting on that object?
I am guessing that there is some force that is equal and opposite of the work done by the shopper, which could be explained by Newton's third law
You're comparing work and force on equal terms. They are different things.

But you are on the right track. There must be some sort of force (or component of which that is doing work) that is equal in magnitude and opposite in direction to the component of the shopper's force that happens to be doing work.

To find out which forces do work and how much, find the component of each force that is parallel to the displacement vector. (For example, if a force is completely perpendicular to the displacement, you can ignore it because it's not doing any work. If a force is at some sort of angle to the displacement vector, you need to find the component of that force which is parallel to the displacement vector to calculate how much work the force does.)
Doc Al
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Jun22-10, 06:08 AM
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Quote Quote by mandy9008 View Post
if it is moving at a constant speed then the net work will be zero, right? I remember my professor saying that in class, but he never explained why. I am guessing that there is some force that is equal and opposite of the work done by the shopper, which could be explained by Newton's third law
There are friction forces acting on the cart; they are just enough to 'cancel' the force due to the shopper. (Be careful to distinguish force from work--don't mix them up.)

It's Newton's 2nd law--not the third--that tells you that the net force must be zero. ΣF = ma. Since the velocity is constant, you know that a = 0, thus ΣF = 0.


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