Test 1 questions

Good questions, though in one or two cases there might have been more to the context of the problem than is apparent here (if so please clarify and I'll be glad to add to my reply). See my notes and let me know if you have further questions.

I will note that in some cases these questions go a little beyond what has been covered--not to the point of being an unreasonable extension of these ideas but far enough that this is often a pretty challenging test. For example after you have been assigned problems in Chapters 3 - 5 you will have a clearer picture of how to deal with forces. I do take this into account when grading this test.

I hope that statement is part of a question with a larger context, because the statement isn't true in all circumstances.

The velocity will increase by the same amount if the acceleration and time interval are the same, since vf = v0 + a `dt.

On the other hand if the acceleration and displacement are the same this will not be the case. In that case the change in velocity would be governed by

vf^2 = v0^2 + 2 a `ds,

and it is the change in squared velocity that is dictated by acceleration and displacement.

If the change in velocity is the same for two different initial velocities, then the change in squared velocity will not be the same (e.g., if velocity changes from 3 m/s to 5 m/s the squared velocity changes by 16 m^2/s^2 from 9 m^2/s^2 to 25 m^2/s^2, whereas a change from 5 m/s to 7 m/s implies a change of 24 m^2/s^2 in the squared velocity. The 2 m/s change in velocity gives two very different changes in squared velocity); and if the change in squared velocities is the same then the change in velocities will not be the same (you can check this out if you wish).

You would get force components using the sin and cosine of the angle; you wouldn't multiply force by angle.

For small angles, however, the force component down the incline is nearly equal to the weight times the slope of the incline (not the angle, the slope, which is rise / run; the slope is equal to the tangent of the angle of the incline).

You also wouldn't use the entire 3.234 N as the weight of the object. Only the 2.695 N weight is on the incline; the .539 N weight is suspended so it contributes the all its weight to the force in the direction of motion. The 2.695 N weight is mostly supported by the incline and only a small portion of that weight acts down the incline. See below for more details.

At 4 degrees the weight vector for the cart is at 266 degrees with a positive x axis directed along the incline, provided you picture the incline as sloping down and toward the left. The problem can be analyzed with this orientation but turns out to be easier if the incline is pictured as sloping down and to the right. This way the direction of motion will be in the positive x direction and the signs are easier to keep track of.

So let the cart be sloped down and to the right, which puts the weight vector at 274 deg.

The weight of the cart therefore has components of 2.695 N cos(274 deg) in the x direction, along the incline, and 2.695 N sin(274 deg) in the y direction, which is perpendicular to the incline.

The weight component in the x direction acts in the direction of motion of the system. The force exerted by gravity on the suspended mass contributes direction to the motion of the system. The sum of these forces is 2.695 N cos(274 deg) + .539 N.

The y component of the weight acts perpendicular to the incline, which tends to bend or compress the incline so that the incline exerts an equal and opposite force, which is called the normal force of magnitude | 2.695 N sin(274 deg) |.

This normal force results in a frictional force of .12 | 2.695 N sin(274 deg) | opposed to motion.

Thus the net force in the direction of motion is 2.695 N cos(274 deg) + .539 N - .12 | 2.695 N sin(274 deg) |.

The system has mass .055 kg + .275 kg = .330 kg, and the acceleration is equal to the net force divided by the mass.

Finding the angle at which the acceleration is zero is very challenging, which is why this is only part of the problem. You should look at everything else before you spend a lot of time on this solution, which is mathematically challenging (but still at the level of precalculus):

If the angle is theta, then instead of 274 deg we will use theta, which we will regard as our unknown. We are looking for the value of theta that makes the net force zero.

The net force will be 2.695 N cos(270 deg + theta) + .539 N - .12 | 2.695 N sin(270 deg + theta) |.

Setting this equal to 0 we get the equation

2.695 N cos(270 deg + theta) + .539 N - .12 | 2.695 N sin(270 deg + theta) | = 0.

The most basic of trigonometric identities tells us that

sin^2(270 deg + theta) + cos^2(270 deg + theta) = 1, so that

| sin(270 deg + theta) | = sqrt( 1 - cos^2(270 deg + theta) ).

We can plug this into the equation. To simplify the form let u = cos(270 deg + theta).

The equation therefore becomes

2.695 u + .539 N - .12 * 2.695 sqrt(1 - u^2) = 0.

We can rearrange to get

.12 * 2.695 sqrt(1 - u^2) = 2.695 u + .539 N

then square both sides, which gives us a quadratic equation (also multiply out the .12 * 2.695 N). Solving by the quadratic formula gives us two values of u.

Setting cos(270 deg + theta) = u, for each value of u, gives us two equations that we can solve to get theta.

It is certainly possible to get a negative acceleration, given the right quantities. When you get a negative acceleration, this means that the acceleration is in the direction opposite the direction you chose as the positive direction.

With an Atwood machine we usually choose counterclockwise as the positive direction.

The net force is the sum of all the forces acting on the object. It isn't directly related to how far the object travels. There might be more to the context of the problem on which this question arose. For example if you know how much work was done by the net force, the you could find the average force from the displacement of the object.