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anonymous

  • one year ago

How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.4 m across a rough floor without acceleration, if the effective coefficient of friction was 0.70?

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  1. anonymous
    • one year ago
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    So work is defined as a force over a distance. We know the distance (10.4 m) so we just need to find the force. Any ideas how we can find the force here?

  2. anonymous
    • one year ago
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    @LifeEngineer m*a? or μ•F??

  3. anonymous
    • one year ago
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    Right so F=m*a is a great equation if to find force if you know the acceleration of the object. In this case, the crate is being moved at constant speed so we can't use that one. Your second equation is promising though! Friction force, which is what we are looking for here, is defined by F_f=μ•F_N where F_N is the normal force and μ is the coefficient of friction. To visualize whats happening here, we can draw a quick free body diagram: |dw:1434934390600:dw|

  4. anonymous
    • one year ago
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    Any unbalanced force will cause an acceleration inversely proportional to the mass (F=m*a or a=F/m). we know that the crate isnt accelerating in the vertical direction (it doesnt fall through the floor or decide to jump off the ground) so we know that the normal force of the ground pushing up on the crate has to balance the weight of crate. We also know that the crate is moved at a constant velocity. That means the acceleration is zero. So once again, the push force has to balance the friction force. That means if we can find the friction force, we know the push force and can find the work done by the person

  5. anonymous
    • one year ago
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    We know that the weight and the normal force have to balance here so: N=W We need N to find the friction force so if we can find the weight, we can find the friction force. Do you know how to find the weight?

  6. anonymous
    • one year ago
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    @LifeEngineer is it mass X gravity?

  7. anonymous
    • one year ago
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    yep :) So W=m*g g is generally approximated to be g=9.81 m/s^2 You know mass, so you can find Weight and subsequently N (because N=W) Once you've got N, you can find friction force using F_f = μ•N And once you find F_f, you can find work using Work = F*x where x is the distance travelled

  8. anonymous
    • one year ago
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    @LifeEngineer So.... 46*9.8= 450.8, which also equals to N? and now the equation for F_f would be .... .70*450.8 making F_f = 315.56. and then i dont understand where to go from there

  9. anonymous
    • one year ago
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    Perfect so far! So as I said earlier, the easiest way to think about work is that its a force done over a distance. If it helps you remember, when you work out, you lift weights against gravity - the force is the weight of whatever you're lifting. The distance is how high you lift the weight. If you lifted 1 N of weight 1 m high, you would have done 1 Nm or 1 Joule of work. In this case you are pushing against friction instead of lifting against gravity, but its the same principle. You are doing work against a resistance over a distance. We know the Friction force (and we already said that it equaled the Push force). So we know how hard the person is pushing, and we were given how far they push the crate. The work is just F*d where F is the push force and d is the distance

  10. anonymous
    • one year ago
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    @LifeEngineer im not sure i understood that either but i tried it out, does it make the answer come out to 1.4? im not even sure what units go after it either

  11. anonymous
    • one year ago
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    So your force is P=F_f = 315.56 N Your distance is d = 10.4 m Work = Force * distance = 315.56 N * 10.4 m = 3281.8 Nm = 3281.8 Joules = 3.281 kJ

  12. anonymous
    • one year ago
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    @LifeEngineer see thats what confused me because the answer that was on my test that i am going over my teacher wrote that it was 3300 J, thats why i was trying to figure out how exactly he came to get that answer so i know for the next time

  13. anonymous
    • one year ago
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    So yeah, if you round to 2 sig figs, you'd get 3300 J (which is technically correct since the friction factor was only given to 2 sig figs)

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