# Fun question!

1. Jun 15, 2004

### 2Pac

160 newton box sits on 10 meter long frictionless plane inclined at an angke of 30 degrees to the horizontal. calculate the amount of work done in moving the box from the bottom to the top of the inclined plane.

2. Jun 15, 2004

### AKG

$$W = \vec{F}_{app} \cdot \Delta \vec{d}$$
$$W = (m \vec{g} \sin \theta ) \cdot \Delta \vec{d}$$
$$W = 800J$$

Also, I don't believe the path matters, so you can treat the problem as simply lifting the 160N box a height equivalent to the heigh of the ramp:

W = 160N x (10m)[sin(30$^{\circ}$)] = 800J

3. Jun 15, 2004

### 2Pac

thank you AKG. What is the deal with this formula. W=FD cos ? because i got 800J as well and everyone in my class said i was wrong.

4. Jun 15, 2004

### AKG

The dot product between two vectors, $\vec{u}$ and $\vec{v}$ is:

$$\vec{v} \cdot \vec{u} = |\vec{v}||\vec{u}|\cos \theta$$

where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$. And since the cosine function is symmetric about the y-axis, whether you measure from vector u to vector v, getting a positive angle, let's say, or measure from v to u, of course then getting a negative angle, it won't matter.

Now, Work is the dot product of the applied force and the direction of motion. The force you'd have to apply to push the block up the incline is obviously in the up-the-incline direction. That is also the direction of motion, so the angle between the two is zero, and cos(0) = 1. So, the work done is the magnitude of the applied force, times the magnitude of the displacement (times 1). The magnitude of displacement is 10m. If you draw the free-body diagram, you'll see that there's a normal force and gravitational force. Part of the gravitational force is counterbalanced by the normal force, and part of it would then have to be counterbalanced by the applied force, so you can calculate this applied force.

But what if you don't want to just counterbalance the applied force? What if you apply a greater force? The force-applied would be different, and you'd think you'd get a bigger result, right? No, because after some time you'd actually have to pull back on the object and apply a force opposite the direction of motion so that it stops at the end of the ramp. Or, maybe you'll only apply a force for the first 5 meters, then, having given it sufficient velocity, it slides to the top where it just stops. But in this case, although the applied force has increased (accelerating the block rather than just pushing it past equilibrium), the distance decreases. So no matter how you do it, if you bring it from rest at the bottom to rest at the top, the work will be the same.

Another way to calculate work is change in energy. At the bottom and top, it is at rest, so the kinetic energy change is zero. The gravitational potential energy changes, which is what I was getting at in the second approach where I said: "Also, I don't believe the path matters, so you can treat the problem as simply lifting the 160N box a height equivalent to the height of the ramp." Of course, I can't tell you why your class says you have it wrong. I'm pretty convinced I have it right...

5. Jun 15, 2004

### 2Pac

After reading that i am too. thank you again.