BR Physics Work/Energy Passage 2 Questions...

[salsasunrise123]

Hello, So passage 2 in work/energy chapter is a roller coaster loop type problem. My first question is with #9. It asks for velocity as object is at top of loop but it says that object is barely hanging on. The question is below:


If the cart is barely able to complete the loop, what is its
speed at point P, the top of the loop?

My second question deals with # 10. I simply do not understand explanation they give:

Q:

The cart is brought to rest because of the frictional force
acting from Y to Z. The work done by friction to stop
the cart under normal conditions is W. Under rainy
conditions, how much work is done by friction to bring

the cart to rest, assuming the cart reaches point Y with

its normal speed and the frictional force is reduced by
half? (Assume the tracks are long enough to bring the

cart to a halt.)

A:

We should use the work-kinetic energy theorem, which says W = KE. Since the cart reaches
point Y with its normal speed, it has its normal kinetic energy. Since it arrives with its normal kinetic energy, it has the same
KE. If it has the same KE as a cart in normal conditions, the work required to stop the cart is the same. It is spread out over a greater distance.



Lastly, they ask for net force halfway up loop when object is on the horizontal. I thought the net force would be the centripetal force directed towards center of loop and not include the force of gravity. Why do you include the force of gravity even though it plays no role in the horizontal/centripetal force. Thanks.

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[Odi]

What exactly is your question regarding #9?

With #10 think of friction as "absorbing" the kinetic energy in the form of work from any moving object. If two objects have the same initial KE, it doesn't matter if object 1 travels 100m and object 2 travels 1m before coming to a stop. Their kinetic energy was equal when they started and thus they dissipated the same amount of energy to friction.

In #10 they are clear in telling you both carts have the same initial KE and are allowed to roll until they come to a stop. Even though the cart on the rainy day travels further because of a weaker frictional force, it had the same amount of KE as the normal cart and thus they both lose the same amount of KE to friction--in the form of work.

For your last question,

Remember that centripetal force is not a unique force in itself. It can be any force: tension in a string (swinging a rock in a circle), force of gravity (orbiting planets) etc. In the case of a cart moving in a loop, the centripetal force IS the normal force acting on the object. They are the same force, not two different forces. You only have N and mg acting on the cart at all times. Mg is always acting downwards on the object regardless of where the cart is but the normal force is always changing direction and pointing in the centre of the loop (because it's the centripetal force). When they ask for the net force, they are asking for the resultant vector force of all forces acting on an object regardless of whether they are vertical or horizontal.

 

[salsasunrise123]

Thanks. I have another question from this chapter. It is question 40 in the 52 question test. The image is of a ramp with object at top that is let go and slides down. The question asks what will happen if you increase theta while keeping the height the same. One of the choices says that speed at bottom will increase, but it is wrong. Book says that although acceleration will increase, distance decreases so therefore the speed doesnt increase. Im confused as to why that is the case. I though that speed at bottom = sqrrt2GH. Increasing theta would increase acceleration which would increase G, thereby increasing final speed. Can someone please explain this to me?

Im struggling with Berkeley Review (9's on first 3 chapters). I am getting only 1 wrong on about half the passages, but on the others I'm getting around 3 wrong per passage. My goal is to consistently get no more than 1 wrong per passage. Any suggestions?

 

[salsasunrise123]

bump

 

[Odi]

Potential energy depends on mgh. Neither m, g, or h are changing as they increase theta and keep height constant.

That is, there is the same amount of initial PE being converted to final KE as the box reaches the bottom in both sceneries... therefore final KE is the same in both sceneries. In [KE=1/2mv^2] neither mass nor speed must change.

Speed is solved for using PE and KE because the v^2 in KE is speed which is a scalar. It's technically not a velocity vector as it has no direction. Whenever you read speed, think initial and final kinetic energy.

On your free body diagram the force causing the acceleration is [mgsin(theta)= ma] and mass cancels so [gsin(theta) = a] ...since theta is increasing, acceleration must be increasing. Acceleration is the rate of change of velocity (vector) over time. It doesn't tell you about instantaneous speed (scalar).

 

[Iatro]

If there is friction in this problem (i.e. the ramp) then I believe the speed would actually increase if you increased theta (reducing normal force for frictional force component per unit distance. Distance isn't change, so work by friction is reduced, aka removing less energy from block system, resulting in an increased speed. Using work-energy theorem, the amount of work you are subtracting for friction decrease, and so deltaKE increases). Extrapolate the situation, you have theta at 90, you are dropping an object vertically and there is no friction ( normal force is perpendicular to force of friction). This is assuming no air resistance. However, no friction, it's just conservation of energy.