How Much Work Does Friction Do on a Crate Sliding Down a Ramp?

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The discussion revolves around calculating the work done by friction on a 70 kg crate sliding down a 4.0-meter ramp inclined at 30 degrees. The crate starts from rest and reaches a speed of 5.0 m/s at the bottom, leading to confusion over the sign of the work done by friction. The initial calculations suggest a positive work of 500 joules, but the correct interpretation indicates that friction does negative work, resulting in -500 joules. The ambiguity in the question highlights the need for clarity regarding the definition of 'work' in this context, emphasizing that friction opposes motion and dissipates energy. Ultimately, understanding the relationship between gravitational potential energy and the work done by friction is crucial for accurate calculations.
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A 70 kg crate starting at rest slides down a rough ramp 4.0 meters long, inclined at 30 degrees above horizontal. At the bottom, it is moving at 5.0 m/s. what was the work done by friction on the crate?

Answers=
A. -500 Joules
B. 500 Joules
C. 460 Joules
D. -460 Joules
E. 0.0 Joules

I got this answer wrong and i was wondering why... so here is my work. i got the height to be 2meters by using sin 30

Mgh - Wf = 1/2MV^2

70 * 9.81 * 2 - wf = .5 * 70 * 5.0^2
875 - wf = 1373.4
wf =1373.4-875
wf = 500 joules

so i got positive 500 joules. So is the answer actually -500 joules since friction forces go up the ramp>
 
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The question is worded poorly, IMO. The proper phrase should be "how much energy is dissipated by friction?" The ambiguity is in the definition of 'work'. Since the frictional force acts to oppose the motion, the frictional force can be thought of as removing some fraction of the total gravitational potential energy available to perform work.

Of course, the energy is removed as heat and plastic deformation of the materials.
 
cbeeson23 said:
Mgh - Wf = 1/2MV^2
Instead, think:
Total Work done by all forces = ΔKE
Wg + Wf = ΔKE
 
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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