Phy 121 - Assignment 6
A. 1. Work = F cos s, where  is the angle between the force and the direction
of motion. The angle from the horizontal (20° or 30°) is irrelevant.
a. Negative. Friction always opposes motion, so  is 180° any time the force is
friction. cos 180° = -1.
b. Zero. “Normal” means perpendicular; it’s called the normal force because
it’s 90° from the direction of motion. cos 90° = 0.
c. Same. F, s and  (180°) are the same in both cases.
2. a. The work done lifting the bricks equals the potential energy they have
after being lifted: mgh = (200 kg)(9.8 m/s2)(30 m) = 58 800 J. (The kinetic
energy gained when they start moving is lost again when they stop at the top.)
P=
=
= 1960 watts
(1960 W)(
) = 2.63 hp
b. The work done accelerating the car equals the kinetic energy it has after
being accelerated: ½mv2 = ½(1400 kg)(26.8 m/s)2 = 502 768 J. (Assuming the
road is level, no potential energy is gained.)
P=
=
= 83 795 watts
(83795 W)(
) = 112 hp
B. 1. Because the force and displacement are perpendicular.
2.
C. 1. a. No. (Because the force is perpendicular to the direction of motion,
making the cosine factor in the definition of work zero.)
b. Heat. (There is friction, primarily with the air, as the pendulum moves.
Friction converts kinetic energy into thermal energy.)
2. (KE + PE)i + W = (KE + PE)f
KEi +
(-fs)
=
mghf
(PEi = 0) (W = f cos180° s) (KEf = 0: at rest)
540,000 J - (250 N)s = (1200 kg)(9.8 m/s2)(s sin4°)
540,000 - 250 s = 820 s
540,000 = 1070 s
s = 540,000/1070 = 505 m (ans.)
D.
E. a. (PE+KE)A + W = (PE+KE)B
(mghA + 0) + (-35 J) = (0 + ½mvB2)
(22 kg)(9.8)(2.06 m) - 35 = ½(22 kg)(v2)
444.1 - 35 = 11 v2
409.1 = 11 v2
37.19 = v2
6.10 m/s = v (ans.)
b. While falling from B to C (taking down as positive):
Δy = vyit + ½ ay t2
1.15 m = 0 + ½ (9.8) t2 (The y component of vB is zero because she is
moving horizontally.)
1.15 = 4.9 t2
.2347 = t2
.48446 = t That is, about .484 s (ans.)
c.
Δx = (6.10 m/s)(.48446 s) = 2.96 m (ans)