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Lost on magnitude

  1. Oct 5, 2004 #1
    A person weighting 0.6kN rides in an elevator that has a downward acceleration of 1.4m/s^2. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the force of the elevator floor on the person? Answer in units of kN.

    --

    A 1317.9 kg car is coasting along a level road of 33 m/s. A constant breaking force is applied,such that the car is stopped in a distance of 63.7 m. What is the magnitude of the breaking force? Answer in units of N.

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    A 12000 kg sailboat experiences an eastward force 28300 N due to the tide pushing its hull while the wind pushes the sails with a force of 75900 N directed toward the northwest (45 degrees westward of North or 45 degrees northward of West). What is the magnitude of the resultant acceleration of the sailboat? Answer in units of m/s^2.

    What is the direction of the boats acceleration? Answer in units of degrees (N of West).


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    I'm not sure what equations to use on these problems. Any help?
     
  2. jcsd
  3. Oct 5, 2004 #2
    there are only 2 equations
     
  4. Oct 5, 2004 #3
    Well what 2 equations are those? I don't know what to use to solve these problems.
     
  5. Oct 5, 2004 #4
    A person weighting 0.6kN rides in an elevator that has a downward acceleration of 1.4m/s^2. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the force of the elevator floor on the person? Answer in units of kN.

    use F=ma

    600 N=m 9.8 is how they arrived at the weight of a person. So the person weighed
    600/9.8=m

    the mass of the person is 61.2 kg

    F=61.2kg*(9.8-1.4)

    F=61.2kg *8.4 m/s sq
    F=514.248 N
    or 0.5 kN


    The second one you need to a by using the formula V^2=V initial^2 + 2a(x1-x0)

    0=V initial^2 +2a(63.7)

    once you find a replace a in the formula F=ma
     
    Last edited: Oct 5, 2004
  6. Oct 5, 2004 #5

    0.5kN is wrong, is there a different way of solving this problem?
     
  7. Oct 6, 2004 #6

    Pyrrhus

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    Homework Helper

    For the first problem use

    Newton's 2nd Law

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = m \vec{a} [/tex]

    You got movement on the y-axis, the forces are the normal and the weight, the acceleration is downward. Use a free body diagram. Momentum's answer algebraically is correct, if you use webassign or similar try combinations of figures.

    For the third problem

    Remember

    [tex] F = ma [/tex]
     
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