*Because "p" does not correlate to any units of mass that I am familiar with (in Metric or English), I will answer this question assuming "p" to be some generic unit of mass that I will not substitute for anything else (at least not until the end). The important points of solving this problem can still be demonstrated without filling in the units of mass that p represents. Another caveat is that you did not specify the direction (relative to the ground) that the object is moving. For this problem I will assume the object is either moving up or down relative to the ground. If the object is moving horizontally relative to the ground, then the force calculated below is valid but the work is equal to zero. If it is some combination of horizontal and vertical motion, then the answer cannot be solved because there is too little information given. *

Let's start with solving for the force because it is the easiest to find and necessary to calculate the work. By Newton's famous Second Law we have the following equation that dictates that the force "F" experienced by a body is equal to its mass "m" times its acceleration "a":

`F=m*a `

Because we know that the mass of the object is equal to 10p, all that remains is to find the acceleration that the object will experience and substitute into the equation above to find the force. Assuming that this object is in the air on Earth on a very still day (not windy), it will experience an acceleration due to gravity of about -9.81 meters per second squared (the minus sign indicates downward acceleration). Now, all of the known information can be used to solve for the force:

`F=10pxx-9.81(m)/(s^2)=-98.1(p*m)/(s^2) `

Now that the force is known, the work can be calculated. There is another very famous equation called the work integral, which defines work "W" as the integral of the force on an object as a function of the distance traveled "F(y)" over the distance traveled by the object "y" (y1 and y2 represent the starting and ending points of the object's path):

`W=int_(y_1)^(y_2)F(y)dy `

However, if the force on the object is constant across the distance traveled, then this equation can be simplified. In this problem the force is constant across the distance traveled because the force of gravity on the object is always constant (equal to the value we calculated earlier). In that case, and assuming that the object is either moving directly up or down from the ground, the equation above can be used to derive this simpler equation:

`W=F*Deltay `

Where delta y is the distance traveled by the object with the force acting on it. We know every variable that we need in this equation to find the work, so the proper values can be substituted:

`W=-98.1(p*m)/(s^2)*20m =-1962(p*m^2)/(s^2)`

Now something to note is that I have assumed the object is moving upward by plugging in a delta y of 20 meters. If the object moved downward, then -20 meters should be entered above and the sign of the work would become positive. This matters because work has units of energy and can be thought of as the energy given to or taken from the object (in this case by the Earth). The way that I have set up this problem, positive work means that the object gained energy, which would result in the object falling faster (which makes sense if the object is moving downward in gravity). Negative work would mean that the object lost energy and it would be rising more slowly by the end of it's 20 meter travel (which is what happens to objects that move upward against gravity). In either case, the magnitude of the work calculated here holds true.

For exact numerical answers, the mass unit "p" needs to be known. If p were equal to kilograms (kg), then finalizing the answers to metric units for force (Newtons) and energy (Joules) would be simple:

`F=-98.1(kg*m)/(s^2)=-98.1N `

`W=-1962(kg*m^2)/(s^2)=-1962J `

If p were equal to something else like the the English pound mass, then you would simply find the conversion factor from pounds to kilograms, substitute p for pounds, and multiply each of the generic answers found earlier by the conversion factor.

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