By Steven Holzner

One of the forces you’re asked to deal with frequently in physics is the force on an object from a gravitational field. Near Earth or another large mass (such as the Moon), all objects experience a downward force from gravity; on the surface of Earth, every kilogram of mass experiences a downward force of 9.8 N. If the only force acting on an object near Earth is gravity, then that object will have an acceleration of 9.8 m/s2 down, which is about 32 ft/s2 down.

Because the equation sigmaF = ma holds for all forces, you can calculate the force an object feels due to gravity. Here are a few things to keep in mind when dealing with force and gravity:

  • The force on an object is proportional to the object’s mass. For example, twice the mass means twice the force.

  • Because a = F/m, twice the force still means the same acceleration if you have twice the mass. The acceleration due to gravity is constant near Earth’s surface, not the force. Both a 1-kg object and a 2-kg object will fall with an acceleration of 9.8 m/s2 downward near Earth’s surface.

  • The acceleration due to gravity points downward, toward the center of Earth.

The acceleration due to gravity at Earth’s surface carries the symbol g. That means sigmaF = ma becomes F = mg when the only force on an object is gravity; as a vector equation, that’s F = mg, where the vector g is pointing downward.

In practical terms, unless you’re dealing with points so far apart that the curvature of the Earth matters, g is just considered as always acting in the same downward direction.

Sample question

  1. You’re trying to lift a suitcase with a mass of 20 kg. How much force must you apply at a minimum to lift the suitcase?

    The correct answer is 200 N.

    To be able to lift the suitcase, you have to overcome the force due to gravity. Calculate it as follows: F = mg = (20 kg)(9.8m/s2) = 196 N. With significant figures, this rounds to 200 N.

Practice questions

  1. You’re holding a basketball in your hands. If this basketball has a mass of 0.8 kg, how much force must you provide to hold it up against gravity?

  2. A ball drops off a cliff. How fast is it going 1.0 second later (neglecting air resistance)?

  3. A ball drops off a cliff. How far has it gone in 2.0 seconds (neglecting air resistance)?

  4. A skydiver jumps out of a plane. How far has he gone 4.0 seconds later (neglecting air resistance)?

  5. You throw a baseball straight up into the air at 60 m/s. How fast is it going 1.0 second later (neglecting forces from the air)?

  6. You throw a physics book straight up into the air at 30 m/s. How fast is it going 4.0 seconds later (neglecting air resistance)?

  7. Holding a ball over the edge of a cliff, you throw it up at 10 m/s. What is its position 5.0 seconds later (neglecting forces from the air)?

  8. Holding a ball over the edge of a cliff, you throw it up at 20 m/s. What is its position 7.0 seconds later (neglecting forces from the air)?

Following are answers to the practice questions:

  1. 8 N

    1. Use the equation F = mg.

    2. The weight of the basketball is F = mg = (0.8)(9.8) = 7.8 N. Rounding for significant figures, you get 8 N.

  2. 9.8 m/s, downward

    1. The acceleration due to gravity is g.

    2. The velocity of the ball is vf = at = gt.

    3. Plug in the numbers: vf = (9.8)(1.0) = 9.8 m/s, downward.

  3. 20 m

    1. The acceleration due to gravity is g.

    2. The distance the ball has gone in a time t is

      image0.jpg

    3. Plug in the numbers:

      image1.jpg

  4. 78 m

    1. The acceleration due to gravity is g.

    2. The distance the skydiver has gone in a time t is

      image2.jpg

    3. Plug in the numbers:

      image3.jpg

  5. 50 m/s

    1. The acceleration due to gravity is g.

    2. Use the equation vf = vo + at = vo + gt.

    3. Plug in the numbers: vf = vo + at = vo + gt = 60 – 9.8(1.0) = 50.2 m/s. With significant figures, the answer rounds to 50 m/s.

  6. –9 m/s

    1. The acceleration due to gravity is g.

    2. Use the equation vf = vo + at = vo + gt.

    3. Plug in the numbers: vf = vo + at = vo + gt = 30 – 9.8(4.0) = –9.2 m/s. With significant figures, the answer rounds to –9 m/s.

  7. –70 m

    1. The acceleration due to gravity is g.

    2. Use the equation

      image4.jpg

    3. Plug in the numbers:

      image5.jpg

      With significant figures, the answer rounds to –70 m.

  8. –100 m

    1. The acceleration due to gravity is g.

    2. Use the equation

      image6.jpg

    3. Plug in the numbers:

      image7.jpg

      With significant figures, the answer rounds to –100 m.