N!
Lv 4
N! asked in Science & MathematicsPhysics · 1 decade ago

Can someone help me with a physics question?

A bowling ball that weighs 70N is dropped from a tower 15m above the ground. Disregarding air friction, what is the kinetic energy of the ball when it reaches the ground?

Could someone explain how to do this question and which formula to use on it?

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  • Anonymous
    1 decade ago
    Favorite Answer

    Answer --> 1,050 J

    How to find it, given the values

    W = 70 N

    d = 15 m

    Vi = 0.0 m/s

    g = 9.81 m/s^2

    We can calculate mass because

    W = m * g

    m = W/g

    m = (70 N) / (9.81 m/s^2)

    m = 7.14 kg

    We don't need it right now, but we'll need it later. Right now, you need to determine the velocity of the ball on impact, through uniform acceleration equations, specifically in this case

    Vf = SQRT { Vi^2 + [2ad] }

    Vf = SQRT { (0.0 m/s)^2 + [ 2 * (9.81 m/s^2) * (15 m) ]

    Vf = SQRT { 294.3 m^2/s^2 }

    Vf = 17.2 m/s

    Since kinetic energy = 0.5 * mass * velocity-squared

    KE = 0.5 * m * v^2

    Now use the mass we found earlier, and the velocity we just found:

    KE = 0.5 * (7.14 kg) * (17.2 m/s)^2

    KE = (3.57 kg) * (294.3 m^2/s^2)

    KE = 1,050 J

    Source(s): Geologist and I did well in college physics
  • 1 decade ago

    The bowling ball has Potential Energy because someone carried it up the tower. It's PE is mgh where mg is the ball's weight and h is given to be 15m. Notice that PE is proportional to the weight and height of the bowling ball: 70N*15m=1050J. The higher or the heavier the ball, the more PE it has. Neglecting air resistance, the PE the ball has 15m above the ground will be equal to it's Kinetic Energy KE at the ground. KE = 1050J The equation for KE is (1/2)mv^2. So, we can write: mgh=(1/2)mv^2 Rearranging: 2gh=v^2 and v = sqrt(2gh) = 17.2m/s

  • 1 decade ago

    First you must figure out how much time it takes to drop all the way. The distance formula you learned in class is

    d = at^2/2

    Solving for time, this becomes

    sqrt[2d/a] = t

    I assume that the acceleration, a, is supposed to be near Earth's surface, a value of 9.81m/s^2. The distance is 15m. Plugging these into the time formula gives

    sqrt[30/9.81] = 1.75 seconds

    The speed of the ball just as it gets to the ground is given by at = v. Plugging in the numbers we have gives us

    9.81*1.75 = 17.17m/s

    Now we must determine the ball's mass. It has a weight of 70N under an acceleration of 9.81m/s^2. Weight is just force, or F = ma. Solving for mass gives us

    F/a = m

    Plugging in the numbers gives us

    70/9.81 = 7.14kg.

    Kinetic energy is given by KE = mv^2/2. Plugging in the values we've determined gives us

    7.14*(17.17^2)/2 = 1,052.47 joules

  • Anonymous
    1 decade ago

    so the formula for kinetic energy is Ek= (1/2)m v^2 where m is mass in and v is velocity in m/s

    so find the mass, fnet= (mass)(acceleration) rearrange so divide the 70N by the accel in this case 9.8 m/s ( the force of gravity) = so the ball weighs 7.1428 kg. Now you have to find the velocity. which is V= (start velocity to power 2)+(2ad) start velocity is 0 m/s, accel is 9.8m/s and dist is 15 m the final velocity is 17. 15 m/s so since EK= 1/2mv^2 (1/2)(7.14)(17.15^2) = 1050 N

    Check my work I haven't done physics since last semester, hope im right

    Source(s): Grade 11 physics
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  • Mitch
    Lv 5
    1 decade ago

    s = (1/2) at^2 (since v0=0)

    15 =(1/2)(9.8)*t^2

    3.06 = t^2

    1.75sec = t

    v = at = 9.8m/sec^2 * 1.75 sec = 17.15 m/sec

    m=F/a = 70N/(9.8m/s^2) = 7.14 kg

    KE = (1/2) mv^2 = (1/2)*7.14*(17.15)^2

    KE = 1050J

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