- Good morning!
Billy could you please read the problem
and Bo, could you please translate?
♫ Flipping Physics ♫
- A shopper pushes a shopping cart with a force
of 6.9 newtons at an angle of 59 degrees
to the left of the negative y-axis.
- Stop.
The force applied is 6.9 newtons at an angle
of 59 degrees to the left of the negative y-axis
- Billy, please continue.
- While the cart moves a horizontal distance of 7.0 meters,
what is the work done by the shopper on the shopping cart?
- Displacement equals 7.0 meters
and work equals question mark.
- I don't really understand the direction,
the 59 degrees to the left
of the negative y-axis.
Could you please explain that?
- Sure Bobby, let's take a look at the example
to better understand what is going on here.
You can see the seven meters
through which the cart travels.
Also, in order to get actual numbers here
for this force, I used a force sensor to measure
that 6.9 newton force.
And as far as the angle is concerned,
you can see the force I am applying
is down and to the left.
And the 59 degree angle which I measured
is between the force applied and the vertical.
Bobby, does that help?
- Yes, thank you.
But doesn't that mean the cart is moving to the left
so the displacement of the cart
should be negative seven meters?
- That is correct, Bobby.
(clattering bicycle spokes)
(dramatic music)
- Two dollars!
(hurried footsteps)
(clattering bicycle spokes)
(clattering cart wheels)
- I want my two dollars!
(dramatic heavy tone)
- We also need a subscript on work
because we always need to identify which force is
or what forces are, doing the work
Billy?
- Well, it asks for the work the shopper does
on the cart and the shopper applies a force
on the cart, so we're trying to determine
the work done by the force applied.
- This problem deals with forces,
Bo, what should we do next?
- Draw the free-body diagram.
- Bo, why don't you do that.
- Ok, the force applied is at an angle of 59 degrees
to the left of the negative y-axis,
the force of gravity is straight down,
the force normal is straight up,
and the force of friction is parallel to the ground
and opposite the direction the cart is moving,
so the force of friction is to the right.
Now we need to break the force applied
into its components, sine--
- Hold up Bo, just wait a minute please.
When I first introduced free-body diagrams
I said we usually break forces into components.
The work equation is an example of when we do not
need to break forces into components, usually,
because the angle is
in the equation for work.
Speaking of the equation for work,
Bobby, could you please give us the work equation?
- Work equals force times displacement
times cosine theta.
- Notice I added subscripts for the force applied
because we are solving for the work done
by the force applied.
Billy, could you please substitute in values?
- The force applied is 6.9
times the displacement which is negative seven,
times the angle, theta,
which is 59 degrees, times the cosine of 59 degrees,
which is negative 24.876 newton meters
or with sig figs, negative 25 joules.
(sharp click)
Ah, wait a minute.
We only used the magnitude of the force and the magnitude
of the displacement in the work equation.
While the displacement is negative seven meters, we actually
need to use positive seven meters in the equation.
- Right, it's the cosine of the angle that determines
if the work is positive or negative.
- Yeah, speaking of the angle,
we used the wrong value.
Theta is the angle between the direction of the force
applied and the displacement of the cart,
which is actually in this case, 90 minus
59 degrees, or 31 degrees.
Which means the work done
by the force applied is actually
41.401
or with two sig figs, 41 newtons.
I'm sorry, 41 joules.
- Glad you all figured that out.
Those are actually two very common mistakes
students make with problems like this.
There's actually more we can do with this problem.
We can figure out the work done on the cart
by the force of gravity, Bo?
- Well, we can substitute in mass times the acceleration
due to gravity for the force of gravity.
However, we don't know the mass of the cart.
Oh! But the force of gravity is straight down
and the displacement of the cart is to the left,
the angle between those two directions is 90 degrees
and the cosine of 90 degrees is zero,
so the work done by the force of gravity is zero.
- Very nice Bo.
We can also figure out the work done on the cart
by the force normal, Bobby?
- Well, the force normal is up
and the cart is displaced to the left.
Again, the angle between those two directions is 90 degrees.
And the cosine of 90 degrees is zero.
The work done by the force normal on the cart is zero,
just like the work done by the force of gravity.
- Thank you Bobby.
Sadly because we don't know the mass,
or the acceleration of the cart,
we cannot figure out the work done on the cart
by the force of friction.
Thank you very much for learning with me today,
I enjoyed learning with you.