Jocelyn C. answered • 06/04/20
Biophysics Major with AP and College Level Teaching Experience
1). The formula for work is W = F•d = Fdcosθ, where F is the force done to displace an object, d is the displacement, and θ is the angle between the force and the displacement. In this case, the total work is calculated using the net force, which is the sum of the frictional force and the pushing force, because both are horizontal:
Fnet horizontal = Fapplied + Ffr = Fapplied + 80 N
(We can assume these are the only horizontal forces on the bed, since no others are mentioned. Because no vertical displacement of the bed occurs, work is only done by the horizontal forces.)
The bed is pushed 3 m, so d = 3. The total work W is given to us as 684 J.
Because the net force we are accounting for is the net horizontal force (see above reasoning) and d is in a horizontal direction, the angle between F and d is 0, and so cosθ = 1. Thus, our work formula is simply W = Fnet horizontald. Solving for Fnet horizontal,
Fnet horizontal = 684/3 = 228 N.
Thus, from our net force equation, Fapplied = 228 - 80 = 248 N.
2). Now, the only horizontal force being applied to our object, the book, is friction: Ffr. Again, we can disregard the vertical forces because no vertical displacement is occurring. Because work can also be calculated as the difference in net energy between two points, we can solve this problem by finding the initial energy of the book right as it leaves the student's hand; its final energy, when the book has stopped/no longer has velocity, is 0 (we set potential energy to be 0 at the elevation of the table, so our net energy calculations only need to consider kinetic energy).
To find the initial energy, we must find the initial velocity of the book as it leaves the student's hand. Note that Ffr is the only force on the book, and it can be assumed to be a constant force. Hence, the acceleration of the book is constant, by F = ma. Since we're given a constant acceleration and the time it takes for the book to stop, we use our constant acceleration kinematics equations.
What we know:
Displacement: Δx = 6.5 m
Time: Δt = 2.25 s
Final velocity: Vf = 0 m/s
What we don't know/need to know:
Initial velocity: V0
Acceleration: a
Given that we're solving for V0, I'm using the linear equations
Vf = V0 + at and Δx = V0t + 0.5at2.
The other equation has us solve for V02 and taking square roots will be annoying, so we use these. Although we're only solving for V0, we need two equations because each of our kinematics equation contains both unknown variables, so we must solve for both.
Plugging in our known quantities:
0 = V0 + 2.25a and 6.5 = 2.25V0 + 0.5·2.252·a.
We solve for V0 in terms of a in the first equation and substitute into the second.
After solving both these equations -- I calculated V0 = 5.778 m/s and a = -2.568 m/s2 -- we calculate initial kinetic energy:
KEi = 0.5mV0 = 0.5(8.5)(5.778) ≈ 141.88 J.
Then, we plug this into our equation for work:
W = ΔE = KEf - KEi = 0 - 141.88J = -141.88J.
Note, however, that we are calculating the work done by friction, which acts in the opposite direction that the book is moving in; hence,
Wfriction = -W = 141.88J.
Our equation for W is really the work done by the book, which loses energy as it slows, which is why Wbook < 0.
Finally, note that we could also have solved for work by solving for acceleration in the kinematics problem, rather than V0, and calculating Ffr = mbooka, then using our original formula, Wfriction = Ffr•d. You would still need to account for the direction of the frictional force, which this time is opposite the direction of motion, so θ = 180º in this case.
Let me know if you find any errors in these solutions!
