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    1) use the masses and the radii of the earth and moon to determine the weight of a man on the surace of then moon if his weight at the earth's surface is 700N.

    mass of earth = 5.98x10^24 kg

    2) Determine the force equired to a pull a 0.500kg vlock at a constant speed of 2.5m/s, if the coefficent of friction is 0.17.

    3) if a 3kg ball has 400J kinetic energy just before before it strikes the ground, then from what height was it released?

    4) calculate the unbalanced force acting on a 5kg cannonball as it accelerates from rest to 150m/s in 0.050s.
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    (Original post by intr3pid)
    2) Determine the force equired to a pull a 0.500kg vlock at a constant speed of 2.5m/s, if the coefficent of friction is 0.17.
    2) because it is at a constant speed there is no resultant force and teh force pulling it is the same as the friction force which is F=uR (u- coefficient) (R=mg)

    F= .17 * (9.81*.5)
    F= .17 * 4.905
    F= 0.83385 N
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    (Original post by intr3pid)
    3) if a 3kg ball has 400J kinetic energy just before before it strikes the ground, then from what height was it released?
    0.5*m*V² = 400
    V² = 400÷(0.5*m) = 400÷1.5
    V² = 266 2/3

    V is the speed at just before collision with ground

    Using V = √(2gh) h is the height from which it was released

    V² = 2gh
    V²÷2g = h
    h= (266 2/3)÷19.6 = 13.61m
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    (Original post by intr3pid)
    4) calculate the unbalanced force acting on a 5kg cannonball as it accelerates from rest to 150m/s in 0.050s.
    I think you have to use impulse stuff on this

    Ft = m(v-u) u=0 (at rest) v=150m/s m=5Kg t=0.05s
    F = (5*(150-0))÷0.05)
    F = 750÷0.05
    F = 15000 N
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    (Original post by intr3pid)
    1) use the masses and the radii of the earth and moon to determine the weight of a man on the surace of then moon if his weight at the earth's surface is 700N.

    mass of earth = 5.98x10^24 kg

    CAN SUM1 PLEEZ DO THIS 1 IM LOST!!!
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    (Original post by mackin boi)
    0.5*m*V² = 400
    V² = 400÷(0.5*m) = 400÷1.5
    V² = 266 2/3

    V is the speed at just before collision with ground

    Using V = √(2gh) h is the height from which it was released

    V² = 2gh
    V²÷2g = h
    h= (266 2/3)÷19.6 = 13.61m
    i would hav done KE gained = PE lost
    so> 400J = m g h
    h = 400/(9.8*3) = 13.61 dont understand what u did :confused: lol
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    (Original post by mackin boi)
    I think you have to use impulse stuff on this

    Ft = m(v-u) u=0 (at rest) v=150m/s m=5Kg t=0.05s
    F = (5*(150-0))÷0.05)
    F = 750÷0.05
    F = 15000 N
    its strange how different ppl do things ...

    i would have done f=ma,
    a = 150/0.05 = 3000
    f = 5 * 3000 = 15000 N

    still we both get the same answer lol

    i not sure bout q1 either ill giv it a go on paper first so i dont take ages tryin to type sodding equations (not easy for the new forum helpers among us)
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    (Original post by mackin boi)
    CAN SUM1 PLEEZ DO THIS 1 IM LOST!!!
    ok here we go: q1

    W on earth = mg 700 = m * 9.81 m = 700/9.81 = 71.4kg (3sf)

    (using mass of moon = 7.36*10^22kg, r of moon = 1,137.4m > from google :eek: )

    g = GM/r^2 >>> g = (6.67*10^-11 * 7.36*10^22)/ (1737.4)^2

    g on moon = 1.63*10^6 (3sf)

    W on moon = m*g = 71.4*(1.63*10^6) = 1.16*10^8N = 116MN.

    dunno if thats right, any clever clogs out there are welcome to show us all the right way.
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    (Original post by Mr Tom)
    its strange how different ppl do things ...

    i would have done f=ma,
    a = 150/0.05 = 3000
    f = 5 * 3000 = 15000 N

    still we both get the same answer lol

    i not sure bout q1 either ill giv it a go on paper first so i dont take ages tryin to type sodding equations (not easy for the new forum helpers among us)
    Tom u r a don...i just like to complicate things
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    (Original post by Mr Tom)
    ok here we go: q1

    W on earth = mg 700 = m * 9.81 m = 700/9.81 = 71.4kg (3sf)

    (using mass of moon = 7.36*10^22kg, r of moon = 1,137.4m > from google :eek: )

    g = GM/r^2 >>> g = (6.67*10^-11 * 7.36*10^22)/ (1737.4)^2

    g on moon = 1.63*10^6 (3sf)

    W on moon = m*g = 71.4*(1.63*10^6) = 1.16*10^8N = 116MN.

    dunno if thats right, any clever clogs out there are welcome to show us all the right way.
    You've made a small (but important!) mistake- the radius is 1,737.4 km, not metres! [check your final answer, it's way too big!]

    you could also just use F = GMm/r² (ie the force the moon exerts on the man)

    F = (6.67*10-11 * 7.36*1022*71.4)/(1737.4*10³)²
    F = 116 N.
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    (Original post by intr3pid)
    1) use the masses and the radii of the earth and moon to determine the weight of a man on the surace of then moon if his weight at the earth's surface is 700N.

    mass of earth = 5.98x10^24 kg

    2) Determine the force equired to a pull a 0.500kg vlock at a constant speed of 2.5m/s, if the coefficent of friction is 0.17.

    3) if a 3kg ball has 400J kinetic energy just before before it strikes the ground, then from what height was it released?

    4) calculate the unbalanced force acting on a 5kg cannonball as it accelerates from rest to 150m/s in 0.050s.
    1. weight at Earth's surface = 700N, mass = 70kg, g on the moon is 1/6 of on Earth, so weight = 700/6 = 117N
    2. resistance = 0.17*0.5*10 = 0.85N,
    3. 400J gpe = mgh, h = 400/mg = 13 m
    4. F=ma = 5*(150/0.05) = 15kN

    all done approximately and in a hurry!
 
 
 
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