Gravity's Work on Mass Sliding Down Incline at 35 Degrees

AI Thread Summary
The discussion centers on calculating the work done by gravity on a 50 kg mass sliding down a 2.0 m incline at a 35-degree angle. The initial calculations yielded a work value of 981, while the teacher's answer was 980, attributed to using g=9.8 instead of g=9.81. Participants suggest applying conservation of energy or the work-energy theorem for further analysis, particularly since friction is negligible in the second question regarding speed. The conversation highlights the importance of precision in calculations and the potential variations in gravitational acceleration based on location. Overall, the thread emphasizes understanding the fundamental physics principles involved in the problem.
._|evo|_.
Messages
13
Reaction score
0

Homework Statement


A 50 kg mass is released from rest and begins to slide from the top of the incline (2.0 m). Th coefficient of kinetic friction between the mass and the inclines u

how much work does gravity do on the mass by the time it slides to the bottom of the ramp if the angle theta is 35 degrees?


Homework Equations


W = f x d

F = m x a

W = (m x a) x d

The Attempt at a Solution



First i found each following value:

m = 50

d = 2.0/sin(35 degrees)

a = 9.81 x sin(35 degrees)

Multiplied these all out, came with 981.

The answer the teacher says is 980 (he's a stickler to exact measurements)

something i did wrong?
 
Last edited:
Physics news on Phys.org
Second question:

What is the speed of the mass by the time it slides to the bottom of the ramp if theta is 35 degrees, and the coefficient of kinetic friction between the mass and the incline is zero?

Related equations:

Um, not sure. Kinda stuck on this part lol.

Attempt:

Can't attempt without a solution.
 
Last edited:
._|evo|_. said:
Multiplied these all out, came with 981.

The answer the teacher says is 980 (he's a stickler to exact measurements)

something i did wrong?

The teacher used g=9.8 instead of g=9.81. Maybe, he lives on an other place on the Earth than you. :wink:


ehild
 
Apply conservation of energy (there is no friction) or the work-energy theorem. You know the work of gravity already.

ehild
 
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