The force, F , between two atoms depends on the distance r separating them. See Figure 1.6. A positive F represents a repulsive force; a negative F represents an attractive force. (a) What happens to the force if the atoms start with r = a and are (i) Pulled slightly further apart? (ii) Pushed slightly closer together? (b) The atoms are said to be in stable equilibrium if the force between them is zero and the atoms tend to return to the equilibrium after a minor disturbance. Does r = a represent a stable equilibrium? Explain. Figure 1.6
The force, F , between two atoms depends on the distance r separating them. See Figure 1.6. A positive F represents a repulsive force; a negative F represents an attractive force. (a) What happens to the force if the atoms start with r = a and are (i) Pulled slightly further apart? (ii) Pushed slightly closer together? (b) The atoms are said to be in stable equilibrium if the force between them is zero and the atoms tend to return to the equilibrium after a minor disturbance. Does r = a represent a stable equilibrium? Explain. Figure 1.6
The force, F, between two atoms depends on the distance r separating them. See Figure 1.6. A positive F represents a repulsive force; a negative F represents an attractive force.
(a) What happens to the force if the atoms start with r = a and are
(i) Pulled slightly further apart?
(ii) Pushed slightly closer together?
(b) The atoms are said to be in stable equilibrium if the force between them is zero and the atoms tend to return to the equilibrium after a minor disturbance. Does r = a represent a stable equilibrium? Explain.
A seasoned parachutist went for a skydiving trip where he performed freefall before deploying
the parachute. According to Newton's Second Law of Motion, there are two forcës acting on
the body of the parachutist, the forces of gravity (F,) and drag force due to air resistance (Fa)
as shown in Figure 1.
Fa = -cv
ITM EUTM FUTM
* UTM TM
Fg= -mg
x(t)
UTM UT
UTM /IM LTM
UTM UTM TUIM
UTM F UT
GROUND
Figure 1: Force acting on body of free-fall
where x(t) is the position of the parachutist from the ground at given time, t is the time of fall
calculated from the start of jump, m is the parachutist's mass, g is the gravitational acceleration,
v is the velocity of the fall and c is the drag coefficient. The equation for the velocity and the
position is given by the equations below:
EUTM PUT
v(t) =
mg
-et/m – 1)
(Eq. 1.1)
x(t) = x(0) –
Where x(0) = 3200 m, m = 79.8 kg, g = 9.81m/s² and c = 6.6 kg/s. It was established that the
critical position to deploy the parachutes is at 762 m from the ground…
It may surprise you to learn that the collision of baseball and bat lasts only about a thou-
sandth of a second. Here we calculate the average force on the bat during this collision by
first computing the change in the ball's momentum.
The momentum p of an object is the product of its mass m and its velocity v, that is,
p = mv. Suppose an object, moving along a straight line, is acted on by a force F = F(t)
that is a continuous function of time.
(a) Show that the change in momentum over a time interval [to, t1] is equal to the integral
of F from to to t1; that is, show that
p(t) – p(to) = |" F(t) dt
This integral is called the impulse of the force over the time interval.
(b) A pitcher throws a 90-mi/h fastball to a batter, who hits a line drive directly back
to the pitcher. The ball is in contact with the bat for 0.001 s and leaves the bat with
velocity 110 mi/h. A baseball weighs 5 oz and, in US Customary units, its mass is
measured in slugs: m = w/g, where g = 32 ft/s².
(i) Find the…
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