A net force of 20 newtons accelerates a four-kilogram object for a distance of six meters. What is the resulting change in the kinetic energy of this object? a) 30 joules, b) 480 joules, c) 120 joules, d) five joules, e) 24 joules.
According to the problem, there’s a net force of 20 newtons. We can use 𝐹 subscript net to represent the net force. This force is used to accelerate an object with a mass of four kilograms. We use the letter 𝑚 to represent the variable of mass. We also know from the problem that our object has traveled a distance of six meters. We will use the letter 𝑑 to represent distance. We are asked to determine the change in the kinetic energy of the object. The Greek letter Δ represents change or difference, and the letters KE represent kinetic energy.
Before we solve our problem mathematically, let’s draw a picture of what’s happening in our problem to make sure that we have a clear idea of what we’re being asked. In our diagram, we have chosen to represent our four-kilogram object as a box. Because the problem did not state a direction for the net force or the distance traveled, we chose the case in which the net force is to the right and the distance traveled was to the right. We represented the motion of our object by starting our four-kilogram box on the left side of our picture, having it moved six meters to the right and completing its acceleration on the right side of the picture.
We have also drawn in our change in kinetic energy to represent the difference in kinetic energy of our initial position and our final position. The fastest way to solve this problem would be to apply the work-energy theorem. Recall that the work-energy theorem states that the net work done on an object is equal to the change in kinetic energy of the object. In equation form, this would be written as 𝑊 equals ΔKE. Recall that we defined work as the force times the displacement. So we also have the equation 𝑊 equals 𝐹𝑑. Combining these two equations together, we get that the net force on our object times the distance our object traveled is equal to the change in kinetic energy of our object.
We can now substitute in the values from the problem into our equation. On the left side of the equation, we replace 𝐹 net with 20 newtons and 𝑑 with six meters. On the right side of the equation, we leave ΔKE the same as that is the variable we’re solving for. Multiplying 20 newtons by six meters, we get 120 joules. The unit of joules comes from one newton times one meter being one joule. Therefore, we can say that the change in kinetic energy of this object is 120 joules.
As a check for answer, we can solve this problem using another method. If we forgot the work-energy theorem, we could still solve the problem using a kinematics-and-forces approach. Recall that the equation for kinetic energy is equal to one-half 𝑚𝑣 squared, with 𝑚 representing the mass and 𝑣 representing the velocity. The change in kinetic energy would be equal to one-half 𝑚𝑣 squared minus one-half 𝑚𝑣 nought squared or kinetic energy final minus kinetic energy initial. 𝑣 represents the final velocity and 𝑣 nought represents the initial velocity.
We can simplify our expression by recognizing that both terms on the right side of the equation have one-half 𝑚 in them. When we pull out one-half 𝑚 from both terms, we’re left with the equation ΔKE equals one-half 𝑚 times quantity 𝑣 squared minus 𝑣 nought squared. When we’re dealing with a change in velocity, we are also dealing with kinematics. Therefore, to determine 𝑣 squared minus 𝑣 nought squared, we must go back and use a kinematics equation. We can use one of our big three kinematics equations that states 𝑣 squared equals 𝑣 nought squared plus two 𝑎𝑑, where 𝑣 is our final velocity, 𝑣 nought is our initial velocity, 𝑎 is our acceleration, and 𝑑 is our distance.
In order to have an expression for 𝑣 squared minus 𝑣 nought squared, we must subtract 𝑣 nought squared from both sides of the equation. 𝑣 nought squared will cancel out from the right side of the equation, leaving the expression 𝑣 squared minus 𝑣 nought squared equals two 𝑎𝑑. We can now substitute in two 𝑎𝑑 for 𝑣 squared minus 𝑣 nought squared in the change in kinetic energy equation. We now have the equation ΔKE equals one-half 𝑚 times two 𝑎𝑑. Since we were not given acceleration in the problem, we must find it. So how do we go about doing that?
Well, we know the net force exerted on the object, and we know the mass. This should remind us of Newton’s second law. Recall that Newton’s second law of motion states that 𝐹 net equals 𝑚𝑎. To isolate the acceleration, we must divide both sides of the equation by mass. This leaves us with the expression for acceleration of net force divided by mass. Substituting into our change in kinetic energy formula, we have ΔKE equals one-half 𝑚 times two times 𝐹 net divided by 𝑚 times 𝑑. Multiplying one-half times two gets us one. Dividing the mass in the numerator by the mass in the denominator also gets us one. This leaves us with the equation ΔKE equals 𝐹 net times 𝑑, which is where we began our calculations when we used our work-energy theorem.
No matter which way we choose to solve the problem, either the shortcut of the work-energy theorem approach or the long way of the kinematics-and-forces approach, we end up with a final answer for the change in kinetic energy of our object to be 120 joules. So the resulting change in the kinetic energy of this object is answer choice c, 120 joules.