Contents
The study of Physics Topics can help us understand and solve real-world problems, from climate change to medical imaging technology.
Is Gravitational Potential Energy Positive or Negative?
Definition: The ability of doing work imparted on a body, by raising it against gravity, is called the gravitational potential energy of the body.
To find the magnitude of gravitational potential energy of a body, a reference plane has to be chosen. The potential energy of the body on the reference plane is taken as zero. The surface of the earth is the usual choice for the reference plane.
Let a body of mass m be raised to a height h from the surface of the earth (reference plane) [Fig.]
Force acting on the body = its weight = mg
Work done against gravitational force
= force × displacement = mg. h
This work done is stored as the potential energy (V) in the body.
∴ V = mgh
or, gravitational potential energy of a body = mass of the body × acceleration due to gravity × height of the body from the reference plane.
In this case, the value of g can be taken as a constant, as the height h is usually negligible compared to the radius of the earth.
Gravitational potential energy is independent of path: Gravitational potential energy depends on the vertical height, but not on the actual path followed to attain that height. Let a body of mass m be raised from the earth’s surface to a height h along a frictionless plane AB of inclination θ [Fig.].
The weight of the body mg acts vertically downwards. Component mg sinθ of this force along the inclined plane acts downwards and the vertical component mgcosθ acts perpendicular to the inclined surface AB. To pull the body up along the inclined surface, work has to be done against the component mgsinθ.
So work done to pull the body from A to B
= force × displacement
= mgsinθ × AB = mg × AB sinθ = mg × BC
[∵ AB sinθ = Bd
= mgh
Hence, the potential energy of the body at a height h above the surface of the earth = nigh.
Therefore, If the height of a body from the reference plane remains the same, the gravitational potential energy also remains the same whatever the path followed to attain that height may be.
Gravitational potential energy depends on reference plane: For measuring the gravitational potential energy generally, the surface of the earth is taken as the zero plane. However, the choice of this plane is totally arbitrary in practice. we do not measure the absolute value of the potential energy, but rather its change. Thus any convenient plane can be taken as the reference plane or zero plane. Therefore, the potential energy will be different, depending on the selection of the reference plane, even if the position of the body remains unchanged.
Gravitational potential energy may be negative:
Gravitational potential energy of a body above the reference plane is positive and of that below the reference plane is negative. For example, when a body of mass m is raised from a point Q on the reference plane. to the point P at a height h1, the work done is mgh1 [Fig.]. This work gets stored as potential energy and hence, the potential energy at the point P is positive. On the other hand, when the mass is taken from Q to R, a point at a depth h2 below the plane of reference, the work done is -mgh2. Hence, the potential energy at the point R is negative.
Potential energy and stable equilibrium: There exists a close relationship between the potential energy and the equilibrium state of a body. An intrinsic property of a body or a system of bodies is to position itself in such a way that, its potential energy is minimum. Equilibrium state of a body is disturbed on increasing its potential energy. Hence, water flows downwards; an extended spring regains its original configuration when released.
Elastic potential energy in an extended spring
Definition: The ability of a body to do work, by virtue of its special shape, is the elastic potential energy of the body.
To extend or to compress a spring, i.e., to give the spring a special shape, work has to be done against the elastic force of the material of the spring. This work remains stored in the spring as its elastic potential energy.
Let x1 = initial length of the spring, and x2 = final length on application of the stretching force.
Hence, extension in length = (x2 – x1) = x (say).
Within the elastic limit, extension of a spring is directly proportional to the stretching force applied.
∴ Applied force = k(x2 – x1), where k is a constant, called the force constant of the spring.
At the initial length x1 of the spring, stretching force on it = O.
Thus, for an extension of (x2 – x1) of the spring i.e., for a displacement of (x2 – x1) of the point of application of the force, an average force of [0 + k(x2 – x1)]/2 or, k(x2 – x1)/2, acts on the spring.
Hence, the work done for the extension of the spring
= average force × displacement
= \(\frac{1}{2}\)k(x2 – x1) × (x2 – x1) = \(\frac{1}{2}\)k(x2……..x1)2 = \(\frac{1}{2}\)kx2
So, the potential energy of the stretched spring = \(\frac{1}{2}\)kx2
For compression also, the potential energy stored in a spring = \(\frac{1}{2}\)kx2, where x is the change in its length.
Deduction using calculus: When a spring is stretched in length by an amount x, an equal and opposite restoring force F is developed in the spring, and F = -kx, where k is the force constant of the spring.
When the length of the spring is further increased by dx, work done against the restoring force,
dW = F ᐧ dx = kxdx [as dx is very small, F is taken as constant over this range]
Hence, the work done against the restoring force for a stretching from 0 to x0,
W = ∫ dW = \(\int_0^{x_0} k x d x\) = \(\frac{1}{2} k x_0^2\)
This work gets stored in the stretched spring as potential energy
Thus, the potential energy of a stretched spring
= \(\frac{1}{2}\) × force constant × (extension)2.
Relation between work and potential energy
Let us consider a stone of mass m, falling freely under gravity. If we ignore air resistance, the net force acting on the stone is its weight mg, which acts downwards. On the other hand, the expression mgh shows that its potential energy increases with h i.e., in the upward direction. In general, the potential energy of an object increases in the direction opposite to that of the resultant force acting on it.
When the stone is lifted from the earth’s surface, the gravitational force and displacement are in opposite directions; so, by convention, the work done by the gravitational force is negative, i.e., W = -mgh. At the same time, the potential energy changes from zero to mgh, i.e, the change in potential energy = mgh. So, we come to the rule that,
work done = -change in P.E.
Numerical Examples
Example 1.
Calculate the kinetic energy of a bullet of mass 50 g moving with a velocity of 200 m ᐧ s-1.
Solution:
Kinetic energy of the bullet = \(\frac{1}{2} m v^2\)
= \(\frac{1}{2}\) × 0.05 × (200)2
= 103 J.
[Here, m = 50g = 0.05 kg, v = 200 m ᐧ s-1]
Example 2.
A ball is moving in air at 15 m ᐧ s-1. The ball is hit with a bat and it attains a velocity of 20 m ᐧ s-1 in the opposite direction. If Its kinetic energy changes by 8.75 J, what is the change In Its momentum?
Solution:
Change in kinetic energy of the ball,
ΔK = \(\frac{1}{2}\)m(\(v_2^2\) – \(v_1^2\))
or, 8.75 = \(\frac{1}{2}\)m(202 – 152) or, m = \(\frac{2 \times 8.75}{35 \times 5}\) = 0.1 kg
∴ The change in momentum of the ball
= m[v1 – (-v2)] = m(v1 + v2)
= 0.1(15 + 20) = 3.5 kg ᐧ m ᐧ s-1.
Example 3.
A body of mass 10 kg falls from a height of 10 m . What will be its kinetic energy just before it touches
the ground? Prove that, this kinetic energy is equal to the potential energy stored at the Initial position of the body. [g = 980 cm ᐧ s-2]
Solution:
Potential energy at the initial position = mgh = 10 × 9.8 × 10 = 980 J.
Let the final velocity of the body just before it touches the ground be v.
So, v2 = u2 + 2gh = 0 + 2 × 9.8 × 10 = 196 m2 ᐧ s-2
∴ Kinetic energy of the body just before it touches ground
= \(\frac{1}{2} m v^2\) = \(\frac{1}{2}\) × 10 × 196 = 980J
Hence, kinetic energy gained = initial potential energy.
Example 4.
A force F acts on a stationary body of mass m for a time t. Show that, the kinetic energy of the body in that time = \(\frac{F^2 t^2}{2 m}\).
Solution:
In this case, force = F; initial speed, u = 0; mass = m.
From F= ma, we can write, a = \(\frac{F}{m}\)
Also, final velocity v = u + at = 0 + \(\frac{F}{m}\)t = \(\frac{F}{m}\)t
∴ Kinetic energy of the body
= \(\frac{1}{2} m v^2\) = \(\frac{1}{2} m\left(\frac{F}{m} t\right)^2\) = \(\frac{F^2 t^2}{2 m}\)
Example 5.
A body of mass m, starting from rest, moves with constant acceleration. After a time T It attains a
velocity V. Show that the work done by the body in time t = \(\frac{1}{2} m \frac{V^2}{T^2} \cdot t^2\). [HS’01]
Solution:
let the velocity attained by the body after a time t be v.
As the acceleration a is constant,
V = aT and v = at = \(\frac{V t}{T}\)
∴ Work done by the body = \(\frac{1}{2} m v^2\) = \(\frac{1}{2} m \frac{V^2 t^2}{T^2}\)
[∵ initial kinetic energy = 0]
Example 6.
A body of mass 1 kg is projected upwards with a velocity of 250 cm ᐧ s-1, from a height 8 m above the earth’s surface. What will be its kinetic energy just before It touches the ground?
Solution:
Let the speed of the body just before it touches the ground be v.
Taking the upward direction as positive,
v2 = u2 + 2gh = (2.5)2 + 2 × 9.8 × 8
= 163.05 m2 ᐧ s-2
[∵ u = 250 cm ᐧ s-1 = 2.5 m ᐧ s-1, g = -9.8 m ᐧ s-2 and h = -8m]
∴ Kinetic energy just before it touches the ground
= \(\frac{1}{2} m v^2\) = \(\frac{1}{2}\) × 1 × 163.05 = 81.525 J.
Example 7.
A body of mass 5 g is moving in a straight line with a velocity of 10 cm ᐧ s-1. A force of 10\(\sqrt{2}\)dyn is applied on the body at an angle of 45° with the line of motion. What is the change in kinetic energy of the body in 1st second? [HS ‘03]
Solution:
Let the line of motion of the body be chosen as the x-axis.
Initial kinetic energy of the body
\(\frac{1}{2} m v^2\) = \(\frac{1}{2}\) × 5 × (10)2 = 250 erg
Components of the applied force along and at right angle to the direction of motion are respectively,
Fx = 10\(\sqrt{2}\)cos45° = 10\(\sqrt{2}\)ᐧ\(\frac{1}{\sqrt{2}}\)dyn
and Fy = 10\(\sqrt{2}\)sin45° = 10\(\sqrt{2}\)ᐧ\(\frac{1}{\sqrt{2}}\)dyn
∴ Components of acceleration of the body,
ax = ay = \(\frac{F_x}{m}\) = \(\frac{F_y}{m}\) = \(\frac{10}{5}\) = 2 cm ᐧ s-1
∴ The velocity along x-axis after 1 s, vx = 10 + 2 × 1
12 cm ᐧ s-1 and along y-axis after 1 s, vy = 0 + 2 × 1 = 2 cm ᐧ s-1
Total kinetic energy of the body after 1 s
= \(\frac{1}{2} m\left(v_x^2+v_y^2\right)\)
= \(\frac{1}{2}\) × 5 × (122 + 22) = \(\frac{1}{2}\) × 5 × 148 = 370 erg
∴ Change in kinetic energy in 1 second
= 370 – 250 = 120 erg.
Example 8.
A bullet of mass 50 g, moving with a velocity of 200 m ᐧ s-1, strikes and penetrates a wooden block. If the resistance of the wooden block on the bullet is 4900 N, find the distance up to which the bullet penetrates the block.
Solution:
Let the distance in the block that the bullet penetrates through be x.
work done = change in kinetic energy
∴ 4900 × x = \(\frac{1}{2}\) × 0.05 × [(200)2 – 02]
∴ x = \(\frac{0.05 \times 4 \times 10^4}{2 \times 4900}\) = 0.204 m.
Example 9.
A bullet, of mass 50 g and of initial speed 400 m ᐧ s-1 penetrates a wall against an average force of 4 × 104N. It comes out with a speed of 50 m ᐧ s-1. What is the thickness of the wall? Another bullet with a lesser mass, but with the same initial velocity, penetrates the wall but is unable to come out. What is the maximum possible mass of the second bullet?
Solution:
In the first case v1 = 400 m ᐧ s-1, v2 = 50 m ᐧ s-1, F = 4 × 104 N, m = 50g = 0.05kg
Let the thickness of the wall be x.
Since, the work done against the force is the change in kinetic energy of the bullet,
Hence, the maximum possible mass of the second bullet is 49.22 g.
Example 10.
A body of mass 0.03 kg, falling from a height of 20 m, penetrates 1.5 m into the soil. Find the average resistive force of the soil.
Solution:
Decrease in potential energy of the body when it comes to rest
= mgh = 0.03 × 9.8 × (20 + 1.5) = 6.32J
Let the average resistive force of soil = F.
∴ F × 1.5 = 6.32 or, F = 4.21 N.