force and wave

1. Use: work done by friction = change of kinetic energy
Ff x 120 = (1/2)m.(20)^2
where m is the mass of the puck, Ff is the frictional force and v is its initial speed
Since Ff = u.(mg), where g is the acceleration due to gravity (taken to be 10 m/s2), and u is the coefficient of friction
hence, umg x 120 = (1/2)m.(20)^2
u = 20^2/(2x120g) = 0.17

2. Since the box is moving with constant speed, the pushing force equals to the frictional force
hence, pushing force = 0.2 x 11.2g N = 22.4 N
where g is the acceleration due to gravity

3. The figures are not clearly seen, I just give you the methods.
Y = A.sin(kx).sin(wt)
The waveform at an instant of time when the vibration is at its maximum amplitude, where sin(wt) =1, is thus given by
y = A.sin(kx)

Nodes are palces with displacement always zero
thus sin(kx) =0
i.e. kx = 0, pi, 2 x pi, 3 x pi, 4 x pi ...etc
where pi = 3.14159...
thus, x = 0, pi/k, 2.pi/k, 3.pi/k ...

Antinodes are just mid-points between two nodes. after you have found the nodes, you can calculate the antinodes accordingly.

4. When the wire is vibrating at fundamental frequency, wavelength = 2 x lingth of wire.
hence wavelength = 2 x 80 cm = 160 cm = 1.6 m
speed of wave = frequency x wavelength = 60 x 1.6 m/s = 96 m/s

since speeed of wave v = square-root[T/m]
where T is the wire tension
m is the mass per unit length of the wire, i.e. m = 40/80 g/cm = 0.5 g/cm = 0.05 kg/m

Thus, 96 = square-root[T/0.05]
solve for T gives T = 461 N

1. By equation of motion: v^2 = u^2 + 2as

0 = (20)^2 + 2a(120)

Acceleration, a = -1.67ms^-2

By Newton's 2nd law of motion, f = ma

Friction, f = -1.67m

Taking the magnitude, f = 1.67m

By the normal reaction, R = mg = 9.8m

So, the coefficient of kinetic friction, u = f/R = 1.67/9.8 = 0.170


2. Frictional force, f = umg = (0.20)(11.2)(9.8) = 21.952 N

As the box is undergoing constant speed, so the horizontal applied force = friction = 22.0 N



1. At the nodes, sinkx = 0

sin0.75pix = 0

0.75pix = npi, where n is an integer,

x = 4n/3 m

For the antinodes, sinkx = 1

sin0.75pix = 1

0.75pix = npi +- pi/2

x = (4n/3 +- 2/3) m

2. At fundamental mode, wavelength, 入 = 2l = 1.6 m

By v = f入

Speed of propagation, v = (60.0)(1.6) = 96 ms^-1

Density, p = m/l = (0.04)/(0.8) = 0.05 kgm^-1

By v = sqrt(T/p)

Tension, T = pv^2 = (0.05)(96)^2 = 460.8 N