The discussion revolves around calculating the distance a brick can travel up a hill before stopping, considering the effects of friction. The problem involves concepts from mechanics, specifically energy conservation and forces acting on an object on an incline.
The discussion is ongoing, with participants providing insights and corrections to each other's calculations. Some have offered guidance on the importance of free body diagrams and the correct application of energy principles. Multiple interpretations of the problem are being explored, particularly regarding the forces acting on the brick and the calculations of work and energy.
Participants mention the complexity of the problem and the various variables involved, indicating a need for clarity on the assumptions made, such as the angle of the incline and the coefficients of friction. There is also a recognition of potential errors in previous calculations, particularly concerning the treatment of gravitational potential energy and work done by friction.
I calculated how much energy in the brick after it was pushed with Kinetic equation and I got 1 J, when the brick stops moving it means the energy was consumed by the friction till 0J, I can find work by W=(delta)E and I got 0-1= -1 JPeroK said:
How did you calculate F? How should you have calculated F?Pao44445 said:
The way to calculate F is to start by calculating the normal force of the 500g block on the ramp. A free body diagram would allow you to clearly see what forces act on that block. Concentrate on the [component of] the forces in the direction perpendicular to the ramp. There are two. The block does not accelerate perpendicular to the ramp. That gives you an easy equation to solve for the normal force.Pao44445 said:
Do you expect the left hand side of that equation to turn out positive or negative?Pao44445 said:
Does that have the right sign for the left hand side?Pao44445 said:
Has the brick gained or lost PE?Pao44445 said:
haruspex said:
That is not correct. There are two other forces acting on the object.Pao44445 said:
You need to correct your calculation of F. It is not given by 0.5 times -9.8. Nor is is clear why you quote cosine in one formula and then sine in the next.
Right you are. And yes, the work done by friction plus the work done by the normal force will be equal to the change in (potential plus kinetic) energy. You can ignore the work done by the force of gravity as long as you are already accounting for it as gravitational potential energy.Pao44445 said:
I read your term (0.5)(-9.8)(dsin25) as being the change in PE, but the sign is wrong.Pao44445 said:
Because of the above, I presumed that had been a typo, a missing 'not'.jbriggs444 said:
Yes, I realized that just after posting, wasn't quick enough in my editing.jbriggs444 said:
That is correct, as long as you bear in mind that the mg sin(25) and mg cos(25) terms replace the mg term. They should not really all be shown in the same diagram.Pao44445 said:
You still have not taken into account my comment about the sign in post #14. If you are measuring d as positive up the slope (which you appear to be doing by putting a cos(180) on the right hand side of the equation) then your expression for the gain in PE is negative. That is obviously not right.Pao44445 said:
Just noticed you have indeed made the error I warned of in post #18. You have counted the work against gravity twice over. You have counted it in the gain in PE on the left of the equation and again as a component of F on the right hand side.Pao44445 said:
μmgcosθ is the frictional force down the slope. You would multiply that by the distance over which the force acted, d. Why would you multiply by cos θ again?Pao44445 said:
haruspex said:
F is the frictional force, ##\mu mg\cos(\theta)##, right? That acts parallel to the slope, downwards. The displacement, d, is parallel to the slope, upwards. So what is the angle between them?Pao44445 said:
180?haruspex said:
Right. So why would you multiply by cos(25o) again?Pao44445 said:
Yes, and you multiply that by mu to get the frictional force. But having done that you now have a force antiparallel to the displacement. So why did you multiply it by cos(25) again?Pao44445 said: