Physics is crazy right?

A man, starting from his home, travels to the mall which is 5 km away to buy veggies for his dinner.

After shopping he returns home, but this time he finds a shortcut which is just 3 km, so he reaches home early. What's the total displacement traversed by the man?

Bonus Question: How much work did the man do? (Consider the frame of reference inside our planet.)

10 km 8 km Cannot be determined 0 km

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Sravanth C.
Feb 29, 2016

We know that displacement is the shortest distance between two points. In this case the man starts from his home, and reaches home again. Physics doesn't care about the path in which the man travelled, it just cares about the shortest distance between the start point and the destination.

Bonus Question: We know that work done is Force times displacement. And in this case the displacement is zero, so: Work = F s = F 0 = 0 J \text{Work}=F\cdot s\\=F\cdot 0=0J

Really physics is crazy!

Great explanation! Just one minor error, though: Work is calculated by the dot product of force and displacement, not the cross product. The cross product gives you a vector as a result and the dot gives you a scalar.

Travis Harrow - 5 years, 3 months ago

Log in to reply

Yes , you are correct. W = F × s W=F\times s , though it is "mathematically correct" , it is totally wrong in physics since work is a scalar quantity. Only dot product fetches a scalar quantity , hence , W = F s W=F\cdot s is the correct equation.

Nihar Mahajan - 5 years, 3 months ago

Thanks! Edited it. ¨ \ddot\smile

Sravanth C. - 5 years, 3 months ago

If we work from the frame outside the earth then the work done can't be 0. Also when we are observing from out of the universe which is expanding we are doing some work always even when we are not showing any movement with respect to earth.

Kushagra Sahni - 5 years, 3 months ago

Log in to reply

Haha nice. I will fix the frame of reference to "inside earth instead". Thanks for the info!

Sravanth C. - 5 years, 3 months ago

The work done is the path integral t r a j e c t o r y F d s \oint_{trajectory}{\vec{F}•\vec{ds}} so we cannot just use the net displacement. Work over a closed loop is generally nonzero when the force field is not conservative (which is usually the case when friction plays a role) and cannot be determined without knowing the force .

K T - 2 years ago

Of course physics is crazy, but you did a great job,keep it up pls.

Muhammad A Abdullahi - 7 months, 3 weeks ago

how did you know that the displacement was zero in the first place? also what do f and J stand for?

Ana Felix Duarte - 2 months, 3 weeks ago

Log in to reply

the displacement is zero sine the person came back to the same point that they started from. F Force J Joules (SI unit for energy) \begin{aligned} F \rightarrow &\textrm{ Force}\\ J \rightarrow &\textrm{ Joules (SI unit for energy)}\end{aligned}

Sravanth C. - 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...