Question 2.8: Equation Jeopardy The first and third equations below are th......
Equation Jeopardy
The first and third equations below are the horizontal x-component form of Newton’s second law and a kinematics equation representing a process. The second equation is for the vertical y-component form of Newton’s second law for that same process:
a_x=\frac{-f_{\mathrm{S} \text { on } \mathrm{O}}}{(50 \mathrm{~kg})}
N_{\mathrm{S} \text { on } \mathrm{O}}-(50 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg})=0
0-(20 \mathrm{~m} / \mathrm{s})^2=2 a_x(+25 \mathrm{~m})
First, determine the values of all unknown quantities in the equations. Then work backward and construct a force diagram and a motion diagram for a system object and invent a process that is consistent with the equations (there are many possibilities). Reverse the problem-solving strategy you used in Example 2.6.
Solve The second equation can be solved for the normal force:
N_{\text {Son } \mathrm{O}}=(50 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg})=490 \mathrm{~N}
We solve the third equation for the acceleration a_x:
a_x=\frac{-(20 \mathrm{~m} / \mathrm{s})^2}{2(25 \mathrm{~m})}=-8.0 \mathrm{~m} / \mathrm{s}^2
Now substitute this result in the first equation to find the magnitude of the force f_{\text {S on } O:}:
-f_{\text {Son } O}=(50 \mathrm{~kg})\left(-8.0 \mathrm{~m} / \mathrm{s}^2\right)=-400 \mathrm{~N}
Represent mathematically The third equation in the problem statement looks like the application of the following kinematics equation to the problem process:
v_x^2-v_{0 x}^2=2 a_x\left(x-x_0\right)
By comparing the above to the third provided equation, we see that the final velocity was v_x=0 , the initial velocity was v_{0 x}=+20 \mathrm{~m} / \mathrm{s} and the displacement of the object was \left(x-x_0\right)=+25 \mathrm{~m} The first equation indicates that there is only one force exerted on the system object in the horizontal x-direction. It has a magnitude of 400 N and points in the negative direction (the same direction as the object’s acceleration), which is opposite the direction of the initial velocity. The second equation indicates that the system object has a 50-kg mass. It could be some kind of a friction force that causes a 50-kg object to slow down and stop while it is traveling 25 m horizontally.
Simplify and diagram We can now construct a motion diagram and a force diagram for the object (see a and b).
Sketch and translate This could have been a 50-kg snowboarder moving at 20 m/s after traveling down a steep hill and then skidding to a stop in 25 m on a horizontal surface (see below). You might have been asked to determine the average friction force that the snow exerted on the snowboarder while he was slowing down.
Try it yourself: Suppose the friction force remained the same but the snowboarder’s initial speed was 10 m/s instead of 20 m/s. In what other ways would the trip be affected?
Answer: The snowboarder would stop in (25 m) / 4 = 6 m.
