We discuss problems, which highlight certain aspects of the study leading to the conservation of energy. The questions are categorized in terms of the characterizing features of the subject matter :

Problem 1: The potential energy of a particle of mass "m" is given by " 1 2 A x 2 1 2 A x 2 ". If non-conservative force is not involved and total mechanical energy is "E", then find its speed at "x".

Solution : External force and non-conservative force are absent here. Hence, we can apply conservation of mechanical energy. Total mechanical energy of the particle is constant,

E = U + K E = U + K

⇒ K = E − U ⇒ K = E − U

⇒ K = E − 1 2 A x 2 ⇒ K = E − 1 2 A x 2

⇒ 1 2 m v 2 = E − 1 2 A x 2 ⇒ 1 2 m v 2 = E − 1 2 A x 2

⇒ v = { 2 E − A x 2 m } ⇒ v = { 2 E − A x 2 m }

Problem 2: A ball of 10 gm is fired from a spring gun of spring constant 500 N/m. The spring is initially compressed by 5 cm and is, then, released. Find the maximum horizontal distance (in meter) traveled by the bullet.

Solution :We are required to find maximum horizontal distance. The horizontal range of a projectile from the ground at an angle “θ” is given by :

Figure 1: A ball is fired from spring gun.

Projectile

R = v 2 sin 2 θ g R = v 2 sin 2 θ g

For maximum horizontal range, the angle of projection should be 45°. The maximum range is :

⇒ R = v 2 g ⇒ R = v 2 g

In order to find speed of projection, we shall employ conservation of mechanical energy. The gun and ball forms an isolated system. Initially, kinetic energy of the bullet is zero, whereas initial elastic potential energy of the spring is “ 1 2 k x 2 1 2 k x 2 ”. Let “v” be the speed when ball breaks the contact with spring. Now, applying conservation of mechanical energy,

K i + U i = K f + U f K i + U i = K f + U f

Here,

K i = 0 K i = 0

U i = 1 2 k x 2 U i = 1 2 k x 2

K f = 1 2 m v 2 K f = 1 2 m v 2

U f = 0 U f = 0

Thus,

⇒ 1 2 k x 2 = 1 2 m v 2 ⇒ 1 2 k x 2 = 1 2 m v 2

⇒ v 2 = k x 2 m ⇒ v 2 = k x 2 m

Putting this expression in the expression of horizontal range, the maximum horizontal range is :

⇒ R = k x 2 m g ⇒ R = k x 2 m g

⇒ R = 500 X 0.05 2 0.01 X 10 = 500 x 0.0025 0.1 = 1.25 0.1 = 12.5 m ⇒ R = 500 X 0.05 2 0.01 X 10 = 500 x 0.0025 0.1 = 1.25 0.1 = 12.5 m

Problem 5: We are designing a delivery ramp for crates containing exercise equipment. The 1470-N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22 degrees. The ramp exerts a 550-N kinetic friction force on each crate, and the maximum static friction also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 8.0 meters along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring which will meet the design criteria. Take sin22° = 0.374 and g = 10 m / s 2 m / s 2 .

Figure 4: The block is moving with an initial speed down the incline.

Incline – block - spring system

Solution : Here the critical part of the question is to understand that the force required to return the crate up is equal to the sum of the component of its weight down the incline and the maximum static friction so that block can be initiated up the incline. Hence, spring force for this condition is :

Figure 5: Forces on the crate.

Incline – block - spring system

F S = m g sin θ + F K = m g sin 22 0 + 550 = 1470 X 0.374 + 550 = 1100 N F S = m g sin θ + F K = m g sin 22 0 + 550 = 1470 X 0.374 + 550 = 1100 N

Let “x” be the compression in the spring. Then, applying Hooke’ law, we get the compression in the spring as :

F S = K x = 1100 F S = K x = 1100

The system here is an isolated system with both conservative and non-conservative force. Hence,

W F = Δ K + Δ U W F = Δ K + Δ U

Here,

⇒ W F = - 550 x 8 = - 4400 J ⇒ W F = - 550 x 8 = - 4400 J

⇒ Δ K = K f − K i = 0 − 1 2 X 147 X 1.82 = - 238.14 J ⇒ Δ K = K f − K i = 0 − 1 2 X 147 X 1.82 = - 238.14 J

⇒ Δ U = 1 2 k x 2 − m g h = 1 2 k x 2 − 1470 X 8 X sin 22 0 ⇒ Δ U = 1 2 k x 2 − m g h = 1 2 k x 2 − 1470 X 8 X sin 22 0

⇒ Δ U = 1 2 k x 2 − 1470 X 8 X sin 22 0 = 1 2 k x 2 − 4398 ⇒ Δ U = 1 2 k x 2 − 1470 X 8 X sin 22 0 = 1 2 k x 2 − 4398

Putting values in the equation, we have :

⇒ W F = Δ K + Δ U ⇒ W F = Δ K + Δ U

⇒ - 4400 = - 238 + 1 2 k x 2 − 4398 ⇒ - 4400 = - 238 + 1 2 k x 2 − 4398

⇒ 1 2 k x 2 = 238 ⇒ 1 2 k x 2 = 238

⇒ k x 2 = 476 ⇒ k x 2 = 476

Combining with Hooke’s equation,

⇒ K = 476 K 2 1100 2 ⇒ K = 476 K 2 1100 2

⇒ K = 1100 2 476 = 2542 N / m ⇒ K = 1100 2 476 = 2542 N / m

Problem 5: An object inside a spherical shell is projected horizontally at the lowest point such that it just completes the circular motion in a vertical plane. If the radius of the shell is 2 m, then find the acceleration of the object when its velocity is vertical.

Figure 6: An object inside a spherical shell is projected horizontally at the lowest point.

Vertical circular motion

Solution : The object attains velocity in vertical direction, when it reaches half of the total height. The acceleration of the object is resultant of tangential and centripetal accelerations at that point :

Figure 7: The acceleration of the object is resultant of tangential and centripetal accelerations.

Vertical circular motion

a = a T 2 + a R 2 a = a T 2 + a R 2

Here,

a T = g a T = g

and

a r = v 2 r a r = v 2 r

We need to know the speed of the object when it has reached half the height. Now, the minimum speed at the lowest point to enable to complete the revolution is :

v L = 5 g r v L = 5 g r

From conservation of energy at lowest and middle levels,

K L + U L = K M + U M K L + U L = K M + U M

Considering zero reference potential at the lowest point,

U L = 0 U L = 0

U M = m g r U M = m g r

Putting these values in energy equation, we have :

⇒ 1 2 m v 2 = 1 2 m 5 g r 2 − m g r ⇒ 1 2 m v 2 = 1 2 m 5 g r 2 − m g r

⇒ v 2 = 5 g r 2 − 2 g r = 3 g r ⇒ v 2 = 5 g r 2 − 2 g r = 3 g r

and

⇒ a r = v 2 r = 3 g r r = 3 g ⇒ a r = v 2 r = 3 g r r = 3 g

Thus,

⇒ a = { g 2 + 3 g 2 } = 10 g = 10 10 m / s 2 ⇒ a = { g 2 + 3 g 2 } = 10 g = 10 10 m / s 2

Note : It is interesting to note that acceleration is independent of the radius of the shell.