Calculating Energy Lost due to Friction

AI Thread Summary
To calculate the work done by a 68.9 N force on a 15.6 kg block dragged over a rough surface, the work is determined to be 297.74 J using the formula W=F(Δr)cos∅. The energy lost due to friction is calculated by first finding the force of kinetic friction, which is affected by the vertical component of the pulling force. The correct normal force is adjusted to account for this vertical component, leading to a revised calculation for kinetic friction. The initial calculation of energy lost due to friction was incorrect at 200.90 J due to not adjusting the normal force properly. Properly accounting for the vertical lift results in the correct determination of energy lost due to friction.
shrutij
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Homework Statement


A 15.6 kg block is dragged over a rough, horizontal surface by a 68.9 N force acting at 19.7 degrees above the horizontal. The block is displaced 4.59 m, and the coefficient of kinetic friction is 0.286. Find the work done by the 68.9 N force.

How much energy is lost due to friction?


Homework Equations


W=F (Δr) cos∅
Friction (kinetic)= μk* m*g


The Attempt at a Solution


I found the work done by the 68.9 N force using the equation above to get 297.74 J.
The energy lost due to friction should be simply the Force of kinetic friction multiplied by the displacement, right?
I did 0.286*15.6 kg* 9.81 to get 43.77 N as the force of kinetic friction.
Energy lost= Force friction * displacement i.e. 43.77 N * 4.59 m = 200.90 J, which is wrong.

What am I missing?
 
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The normal force is reduced by the "lift" of the vertical component of the 68.9 N pulling force.
 
So, normal force should be mg-68.9sin19.7 ?
 
Yes, right on.
 
Got it. Thanks so much!
 
Most welcome! Good luck on the next one.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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