I have read some questions regarding just Work and K.E in these Q&As: Relation and difference between work and kinetic energy AND Understanding relationship between work and energy.

My question includes the involvement of Friction as well.

Assume there is an object being pulled at an angle with the horizontal, with a force, for a certain distance along surface that has friction as well.

By drawing the free body diagram, I get sum of forces in the x-axis to be $$F(net) = (Fcosθ - μmg)$$ because Fcosθ is in the +ve x direction and friction (μmg) is acting in -ve x direction.

We know that: $$W = Fdcosθ$$ Which means, $$W = (Fcosθ - μmg) . d$$

Would this work done, be equal to the K.E of the box after being pulled for the distance d?

Since Energy is conserved when there is no friction or anything that disrupts the system, if accounting for Work done includes the repelling force of friction, this Work done should equal the K.E of the body.

Am I right in understanding the concept?


Yes, you are right, because the theorem states that

$\Delta E_k=\Delta W$

on your system.

  • $\begingroup$ When I check my answer, it says I am wrong. To put it better; F = 611 ; m = 69.9 ; θ = 36.3 ; μ = 0.139 ; d = 5.62 Plugging all that gives me 2230 J (Keeping sig digs in mind) and my portal says it is wrong. $\endgroup$ – Imtiaz Raqib Oct 10 '17 at 3:32
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    $\begingroup$ Nice fail. Well, the theorem above is right and the reasoning is primarily true. The thing is that the normal force is not that one. If you're pulling with a force at angle $\theta$, the force giving work to the $E_k$ is $F\cdot \cos \theta$ but the rest is $F\cdot\sin \theta$. This force acts in the vertical direction and it partially compensates the weight, so the normal force is not $mg$ but $mg-F\cdot \sin \theta$. $\endgroup$ – FGSUZ Oct 10 '17 at 13:04
  • $\begingroup$ I realized that last night and fixed it. Thank you so much for your help! :D $\endgroup$ – Imtiaz Raqib Oct 10 '17 at 16:45

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