Problem#1
A baseball outfielder throws a 0.150-kg baseball at a speed of 40.0 m/s and an initial angle of 30.0°. What is the kinetic energy of the baseball at the highest point of its trajectory?Answer:
At start, v = vxii + vyij
With vxi = vi cosθ0 = (40.0 m/s) cos30.00 = 34.6 m/s and vyi = (40.0 m/s) sin30.00 = 20.0 m/s,
vi = (34.6 m/s)i + (20.0 m/s)j
the baseball at the highest point of its trajectory, vyf = 0 and vxf = vxi then
vf = vxfi + vyfj
vf = (34.6 m/s)i + (0)j
so that
K = ½ mvf2 = ½ (0.150 kg)(34.6 m/s)2 = 90.0 J
Problem#2
While running, a person dissipates about 0.600 J of mechanical energy per step per kilogram of body mass. If a 60.0-kg runner dissipates a power of 70.0 W during a race, how fast is the person running? Assume a running step is 1.50 m long.
Answer:
Concentration of Energy output = (0.600 J/kg.step)(60.0 kg)(1 step/1.50 m) = 24.0 J/m
Then, the speed of the person who runs is
P = F.v
v = 70.0W/24.0 N = 2.92 m/s
Problem#3
The direction of any vector A in three-dimensional space can be specified by giving the angles α, β, and γ that the vector makes with the x, y, and z axes, respectively. If A = Axi + Ayj + Azk, (a) find expressions for cos α, cos β, and cos γ (these are known as direction cosines), and (b) show that these angles satisfy the relation cos2α + cos2β + cos2γ = 1. (Hint: Take the scalar product of A with i, j, and k separately.)
Answer:
(a) A.i = (A)(1) cos α. But also, A.i = Ax.
Thus, (A)(1) cos α = Ax
Then cosα = Ax/A
Similarly, cos β = Ay/A and cosγ = Az/A
(b) show that these angles satisfy the relation cos2α + cos2β + cos2γ = 1,
cos2α + cos2β + cos2γ = (Ax/A)2 + (Ay/A)2 + (Az/A)2
cos2α + cos2β + cos2γ = (Ax2 + Ay2 + Az2)/A2 = 1
Where A2 = Ax2 + Ay2 + Az2
Problem#4
A 4.00-kg particle moves along the x axis. Its position varies with time according to x = t + 2.0t3, where x is in meters and t is in seconds. Find (a) the kinetic energy at any time t, (b) the acceleration of the particle and the force acting on it at time t, (c) the power being delivered to the particle at time t, and (d) the work done on the particle in the interval t = 0 to t = 2.00 s.
Answer:
(a) the kinetic energy at any time t,
K = ½ mv2
With v = dx/dt = d(t + 2.0t3)/dt = 1 + 6.0t2, then
K = ½ (4.00 kg)(1 + 6.0t2)2
K = (2.00 + 24.0t2 + 72.0t4)J
(b) the acceleration of the particle and the force acting on it at time t, we get
a = dv/dt = d(1 + 6.0t2)/dt
a = (12.0t) m/s2 and
F = ma = (4.00 kg)(12.0t) m/s2
F = (48.0t) N
(c) the power being delivered to the particle at time t is
P = Fv = (48.0t)(1 + 6.0t2)
P = (48.0t + 288t3) W
(d) the work done on the particle in the interval t = 0 to t = 2.00 s. We use
W = ∫Pdt
W = ∫02.00s(48.0t + 288t3)dt = (24.0t2 + 72.0t4)│02.00s
W = 1.25 kJ
Problem#5
The spring constant of an automotive suspension spring increases with increasing load due to a spring coil that is widest at the bottom, smoothly tapering to a smaller diameter near the top. The result is a softer ride on normal road surfaces from the narrower coils, but the car does not bottom out on bumps because when the upper coils col lapse, they leave the stiffer coils near the bottom to absorb the load. For a tapered spiral spring that compresses 12.9 cm with a 1 000-N load and 31.5 cm with a 5 000-N load, (a) evaluate the constants a and b in the empirical equation F = axb and (b) find the work needed to compress the spring 25.0 cm.
Answer:
(a) We write F = axb,
When F = 1000 N and x = 12.9 cm, then
1000 N = a(12.9 cm)b (*)
When F = 5000 N and x = 31.5 cm, then
5000 N = a(31.5 cm)b (**)
we divide the equations (*) and (**), we get
5 = (31.5/12.9)b
ln 5 = bln(2.44)
b = 1.80
so, that
1000 N = a(12.9 cm)(1.80)
1000 N = a(0.129 m)(1.80)
a = 4.01 x 104 N/m1.8
(b) the work needed to compress the spring 25.0 cm.
W = ∫Fdx
With F = (4.01 x 104 N/m1.8)x1.8, then
W = ∫00.25 m(4.01 x 104 N/m1.8)x1.8dx
= [(4.01 x 104 N/m1.8)x2.8/2.8]│00.25 m
W = (4.01 x 104 N/m1.8)(0.25 m)2.8/2.8
W = 294 J
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