A question about work and energy

AI Thread Summary
Two observers moving relative to each other will perceive different velocities of a block, affecting its kinetic energy. While both observers will agree on the acceleration of the block, they will not measure the same displacement due to their relative motion. The work done by friction and total work done on the block cannot be the same for both observers because they measure different distances. The discussion highlights the importance of specifying the relative motion of the observers in physics problems. Ultimately, only the acceleration of the block remains consistent across both observers.
Prem1998
Messages
148
Reaction score
13

Homework Statement


Consider two observers moving with respect to each other at a speed v along a straight line. They observer a block of mass m moving a distance x on a rough surface. The following quantity (or quantities) will be same as observed by the two observers:
(a) kinetic energy of the block at time t
(b) work done by friction
(c) total work done on the block
(d) acceleration of the block

Homework Equations


This is only a thoughtful question so I don't think equations are needed. But if it is a rule, then:
KE=1/2mv^2 W= integration Fdx or FX when F is constant, etc.

The Attempt at a Solution


I figured out that the answers should be b,c and d. It can't definitely be 'a' because the velocity of the block seen by both observers will be different. Now, since none of the observers are accelerating themselves, so they both will see the same acceleration of the block, so 'd' is correct. It is not a relativistic physics question, so 'm' will be the same for both. So, resultant force F=ma will also be same for both. And, distance moved, x is also same for both. So, total work done should be same, hence, 'c' is correct. Again, for both observers, the normal reaction on the block is same and coefficient of kinetic friction is also same. So, both observers will observe the same frictional force on the block and again the same distance moved 'x'. So, work done by friction is also same. Hence, b is correct.
But, in the answers section in the book, it was written that only 'd' is correct. Doesn't assuming that 'd' is correct lead to the conclusion that 'b' and 'd' are also correct?
 
Physics news on Phys.org
Prem1998 said:
And, distance moved, x is also same for both.

Why do you think that distance moved would be same for both the observers ?
 
conscience said:
Why do you think that distance moved would be same for both the observers ?
Why would the distance 'x' change? There is negligible length contraction here.
 
If the two observers are moving with respect to each other ,they cannot measure same displacement of the block .
 
conscience said:
If the two observers are moving with respect to each other ,they cannot measure same displacement of the block .
Oh, I'm sorry. So, that eliminates 'b' and 'c' even if the resultant and frictional forces are same for both observers. Thanks. Isn't the question incomplete as it did not say in the book that v=/=0. I mean, v could be 0 leading to the same value of 'x'.
 
Prem1998 said:
Isn't the question incomplete as it did not say in the book that v=/=0. I mean, v could be 0 leading to the same value of 'x'.

The question clearly states that the two observers are moving relative to each other
That rules out the possibility of them being at rest .

Instead if the question had only stated that there are two moving observers , then there could have been a possibility that the two observers were at rest with respect to each other measuring same displacement of the block .
 
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 hanged 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