Problem#1
Consider an object on which the net force is a resistive force proportional to the square of its speed. For example, assume that the resistive force acting on a speed skater is f = –kmv2, where k is a constant and m is the skater’s mass. The skater crosses the finish line of a straight-line race with speed v0 and then slows down by coasting on his skates. Show that the skater’s speed at any time t after crossing the finish line is v(t) = v0/(1 + ktv0). This problem also provides the background for the two following problems.
Answer:
We use Newton's second law
∑F = ma
–kmv2 = mdv/dt
–kdt = dv/v2
–k∫0tdt = ∫v0v v-2 dv
–kt = 1/v0 – 1/v
1/v = 1/v0 + kt
1/v = (1 + v0kt)/v0
v = v0/(1 + v0kt)
Problem#2
(a) Use the result of Problem 40 to find the position x as a function of time for an object of mass m, located at x = 0 and moving with velocity v0i at time t = 0 and thereafter experiencing a net force –kmv2i. (b) Find the object’s velocity as a function of position.
Answer:
(a) From problem#1,
v = dx/dt
v0/(1 + v0kt) = dx/dt
∫0xdx = ∫0t[v0/(1 + v0kt)] dt
x = k-1∫0t(1 + v0kt)-1kv0dt
x = k-1ln(1 + v0kt)l0t
x = k-1ln(1 + v0kt)
(b) we have
ln(1 + v0kt) = kx
(1 + v0kt) = ekx
or
v0/v = ekx
v = v0e-kx
Problem#3
At major league baseball games it is commonplace to flash on the scoreboard a speed for each pitch. This speed is determined with a radar gun aimed by an operator positioned behind home plate. The gun uses the Doppler shift of microwaves reflected from the baseball, as we will study in Chapter 39. The gun determines the speed at some particular point on the baseball’s path, depending on when
the operator pulls the trigger. Because the ball is subject to a drag force due to air, it slows as it travels 18.3 m toward the plate. Use the result of Problem 41(b) to find how much its speed decreases. Suppose the ball leaves the pitcher’s hand at 90.0 mi/h = 40.2 m/s. Ignore its vertical
motion. Use data on baseballs from Example 6.13 to determine the speed of the pitch when it crosses the plate.
Answer:
We write relationship
– ½ DρAv2 = –kmv2
so
k = DρA/2m
k = (0.305)(1.20 kg/m3)(4.2 x 10-3 m2)/[2(0.145 kg)]
k = 5.3 x 10-3/m
then
v = v0e-kx
v = (40.2 m/s)e-(0.0053/m)(18.3 m) = 36.5 m/s
Problem#4
You can feel a force of air drag on your hand if you stretch your arm out of the open window of a speeding car. [Note: Do not endanger yourself.] What is the order of magnitude of this force? In your solution state the quantities you measure or estimate and their values.
Answer:
We estimate that
D = 1.00, ρ = 1.20 kg/m3, A = (0.100 m)(0.160 m) = 1.60 x 10-3 m2 and v = 27.0 m/s.
The resistance force is then
R = ½ DρAv2
R = ½ (1.00)(1.20 kg/m3)(1.60 x 10-3 m2)(27.0 m/s)2
R = 7.00 N
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