The laws related to work and energy allow us to calculate many interesting and useful things about the world around us. Let's start with a basic work problem, but let's solve for the average applied force assuming we already know the amount of work done.

1. If 256 J of work is required to lift an object against the force of gravity to a height of 4.20 m above the ground in a time of 6.50 s,

a. how much force must be applied to lift the object?
b. how much power is required?
c. what is the weight of the object?

a. Using the work equation, and solving it for the force,

W = F dbecomes F = W / d

F = W / d = 256 J / 4.20 m = 61.0 N

This is the average force applied to lift the object against the force of gravity.

Remember the unit J = N-m so the meters will cancel leaving just newtons.

b. Now we need to find the power.

P = W / t = 256 J / 6.50 s = 39.4 W

c. To determine the weight of the object, we must look back in Chapter 3 and apply Newton's third law.

F₁ = - F₂, where F₁ is the object's weight and F₂ is the applied force from part a.

w = F₁ = - F₂ = - 61.0 N

The negative sign indicates that the force is downward, toward the center of the Earth. Thus, g will also be negative, because the acceleration due to gravity is also downward.

Remember that the mass of the object is:

mass = w / g = -61.0 N / -9.80 m/s² = 6.22 kg

Note: the unit N = kg-m/s² so the m/s² cancels, leaving only kg.

We can then check our answers in problem 1 by using the equation for work done against gravity:

W = m g h = (6.22 kg)(9.80 m/s²)(4.20 m) = 256 kg m²/s² = 256 N-m = 256 J

This is the same amount of work that we were originally given in the problem, so our solution must be correct. Note that the units "kg m²/s²" are the same as "joule." If we had been asked to find the gravitational potential energy of the object in problem 1, we would have used the same setup for the problem because the work done is equal to the change in potential energy, and if we start from a position where h_I = 0,

W = E_p = m g h_F - m g h_I = (6.22 kg)(9.80 m/s²)(4.20 m) - 0 = 256 J

The other type of energy we can calculate using the material in Chapter 4 is the kinetic energy. Kinetic energy is the energy of motion, so the object under study will have a changing velocity if work is being done on it. Let's determine the work needed to change the motion of a car from a constant velocity to a state of rest; that is, to bring a moving car to a stop.

2. How much work is needed to bring a car to a complete stop in 7.0 s if the car has a mass of 1200 kg and is initially moving with a velocity of 37 m/s?

The work done on the car, by friction between the tires and the road, must be equal to the change in kinetic energy of the car.

W = E_k = ¹/₂ m v_F² - ¹/₂ m v_I²

= 0 - ¹/₂ (1200 kg) (37 m/s)² = - 8.2 x 10⁵ kg m²/s² =
- 8.2 x 10⁵ J

Do you see why the final kinetic energy is zero? At rest the car has a velocity of zero, so the kinetic energy must also equal zero.

By now you should be able to figure out what the "-" means. It says that the car is slowing down, so work must be done on the car to bring it to a stop.

If the car comes to a stop in 7.0 seconds, we can find the power needed.

P = W / t = 8.2 x 10⁵ J / 7.0 s = 1.2 x 10⁵ W

We can even find the number of horsepower required.

P = 1.2 x 10⁵ W (1 hp / 746 W) = 160 hp

So far we have looked at cases where some external work has been done on the system, so energy has not been conserved. Let's consider a case where no external work is done and we can use the conservation of energy law to solve the problem.

3. A 5.0-kg ball is at rest 150 meters above the ground before it is dropped. If we neglect air resistance, find the speed of the ball just before it strikes the ground.

We can set the potential energy at the top equal to the kinetic energy at the bottom of the fall because of the conservation of energy law. Remember, it is the total energy at the top and bottom that are equal to each other, but since the ball is at rest (no velocity) at the top, the ball's kinetic energy at the top is zero. At the bottom the potential energy is zero because the height above the ground can be considered to be zero at this level. Therefore we can write:

E_{Total at the top} = E_{Total at the bottom}

E_p(top) + E_k(top) = E_p(bottom) + E_k(bottom)

E_p(top) + 0 = 0 + E_k(bottom)

m g h_top = ¹/₂ m v_F²_bottom

v_F² = 2 m g h / m = 2 (5.0 kg)(9.80 m/s)(150 m) / 5.0 kg

Notice that the masses cancel out in this calculation so we could have done the problem without even knowing what the mass was, but we can also carry the mass values through the calculation as we have shown above.

v_F ² = 14700 kg m²/s² / 5.0 kg = 2940 m²/s²

We are still not quite finished. We must take the square root of both sides of the equation to find v_F and give us the final value for the speed just before the ball hits the ground.

v_F = 54 m/s (downward)

The total energy of an object is not always purely kinetic or simply potential. Total mechanical energy is the sum of the two.

4. How much total energy is possessed by a 3.0-kg cat that is falling at a speed of 2.0 m/s and is still 0.50 m above the floor?

Our cat has both energy of motion and energy of position so:

ET	=	Ep + Ek = m g h + 1/2 m v 2
	=	3.0 kg (9.80 m/s2) (0.50 m) + 1/2 (3.0 kg) (2.0 m/s)2
	=	15 kg m2/s2 + 6.0 kg m2/s2 = 21 J

Don't worry, I really do like cats and this one will land gently on her feet. It is possible to find her final speed just before she lands safely on the floor by using the law of conservation of total energy. Her total energy of 21 J must all be converted into kinetic energy by the time she reaches floor level because her potential energy decreases gradually to zero as she falls. Just before she reaches the floor her kinetic energy will be:

E_k = ¹/₂ m v_F ² = E_T (from above) = 21 J

Solving for speed (v_F) as we did in problem 3 gives us:

v_F = 3.7 m/s (downward)

The applications of work and energy are sometimes not easy to follow and the solution to this type of problem may be somewhat difficult until you have had a little practice. Remember, there are more sample problems in the textbook that you may also work through to get additional practice using these concepts.

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