Work and energy with normal force

AI Thread Summary
To find the work done by the normal force on a block sliding down a rough incline, it's essential to recognize that the normal force acts perpendicular to the direction of motion. Since work is calculated using the formula Fcos(theta)d, and the angle for the normal force is 90 degrees to the slope, the cosine of 90 degrees is zero, indicating that the work done by the normal force is indeed zero. However, the frictional force, which is influenced by the normal force and the coefficient of kinetic friction, does do work against the motion of the block. Understanding these forces is crucial in solving problems related to work and energy on inclined planes. The discussion emphasizes the importance of analyzing force directions to determine work done in physics problems.
Maiia
Messages
78
Reaction score
0

Homework Statement


Starting from rest, a 5.85 kg block slides 2.07m down a rough 26.4degree incline. The coefficient of kinetic friction between the block and the incline is 0.427. The acceleration of gravity is 9.8m/s^2. Find the work done by the normal force. Answer in units of Joules.

Here's my FBD.
physics.jpg


I know the formula for work is Fcos(theta)d. But I don't know how to find the angle for the normal force...
If someone could help out, I would really appreciate it.
 
Physics news on Phys.org
The normal force acts at right angles to the slope.
You 26.4 is in the wrong place - a slope angle of 26.4 means the angle between the slope and horizontal.
 
yes, but if you draw some parallel lines, the 26.4 degrees of the slope angle is also equivalent to that angle up there. oh but i see what you said about the normal being perpendicular...wouldnt that mean work is zero then?
 
Maiia said:
yes, but if you draw some parallel lines, the 26.4 degrees of the slope angle is also equivalent to that angle up there. oh but i see what you said about the normal being perpendicular...wouldnt that mean work is zero then?

Yes you are right that forces in the direction of motion are the ones that do work, but in the case of the Normal force there is a resulting force along the direction of motion as determined by the Normal force * the Coefficient of Friction.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top