AP Physics B & C- Additional Practice Questions (MCQ) on Work, Energy & Power

This post has been a bit delayed because of a ‘boot.com’ virus infection in my system. Fortunately I could disinfect the system without much delay.

A few multiple choice questions on work energy and power were discussed in the last post. Here are a few more multiple choice questions in this section:

(1) A particle of mass 2 kg moving along the positive x-direction has a constant momentum of 6 kg ms^–1. If a constant force of 2 N is applied on the particle for 1 s along the negative x-direction, the change in the kinetic energy of the particle is

(a) decrease of 4 J

(b) increase of 4 J

(c) decrease of 8 J

(d) increase of 4 J

(e) decrease of 9 J

The initial kinetic energy of the particle = p²/2m = 6²/(2×2) = 9 J.

The force F of 2 N actng on the particle for the time t equal to 2 s imparts a momentum Ft equal to 4 kg ms^–1 along the negative direction. Since the initial momentum of the particle is along the positive x-direction, the resultant momentum is 6 – 4 = 2 kg ms^–1.

Therefore, the final kinetic energy of the particle = 1²/(2×2) = 1 J.

The change in the kinetic energy of the particle is 1 – 9 = – 8 J

The kinetic energy of the particle is thus decreased by 8 J.

(2) An elevator motor creates a tension of 5000 N in a hoisting cable and reels it at 0.8 ms^–1. If the efficiency of the elevator system is 80%, the power input to the motor is

(a) 2 kW

(b) 4 kW

(c) 6 kW

(d) 8kW

(e) 10 kW

The power output (P_out) of the system is Fv = 5000×0.8 = 4000 W.

If the power input is P_in we have

efficiency h = P_out/ P_in

Therefore, 80/100 = 4000/P_in so that P_in = 5000 W = 5 kW.

(3) A liquid of density ρ is being continuously pumped through a pipe of area of cross section a. If the speed of the liquid through the pipe is v, the time rate at which kinetic energy is being imparted to the liquid is

(a) av²ρ/2

(b) av³ρ/2

(c) avρ/2

(d) av³ρ

(e) av²ρ

The mass of liquid flowing per second is avρ. Therefore, the time rate at which kinetic energy is imparted to the liquid is ½ (avρ)v² = av³ρ/2

The above questions will be beneficial for AP Physics B as well as C. The following questions are for AP Physics C aspirants:

(4) A machine delivering constant power moves an object along a straight line. The displacement of the object in time t is proportional to

(a) t^–2

(b) t³

(c) t^1/2

(d) t^3/2

(e) t^–3

Since the power is the product of force and velocity we have

Fv = K where K is a constant.

If m is the mas and v is the velocity of the object, the above equation can be written as

M(dv/dt)v = K so that vdv = (K/m)dt

Intrgrating, v²/2 = (K/m)t

Therefore v α t^1/2.

If ‘s’ is the displacement, v = ds/dt so that (ds/dt) α t^1/2.

Integrating, s α t^3/2 [Option (d)].

(5) The velocity v, momentum p and the kinetic energy E of a particle are related as

(a) p = dv/dE

(b) p = dE/dv

(c) p = (dE/dv)^1/2

(d) p = d²E/dv²

(e) p = (dE/dv)²

If the mass of the particle is m we have E = ½ mv²

Differentiating, dE/dv = mv = p.

AP Physics B & C- Work, Energy & Power- Multiple Choice Practice Questions

The essential points you have to remember to answer multiple choice questions on work, energy and power were discussed in the post dated 14^th November 2008. Today we will discuss some multiple choice questions in this section.

The following questions are meant for AP Physics B & C aspirants

(1) A small doll suspended inside a truck using a string of length ℓ, undergoes an angular displacement θ when the truck accidentally collides against a wall and stops. If the truck was moving with uniform velocity towards the wall before the collision, the magnitude of the velocity of the truck was

(a) [2gℓ(1– sinθ)]^1/2

(b) 2gℓ(1– sinθ)]

(c) [2gℓ(1– cosθ)]^1/2

(d) 2gℓ(1– cosθ)

(e) [2gℓcosθ)]^1/2

The kinetic energy of the doll (which serves as the bob of a simple pendulum) is converted into gravitational potential energy when the truck stops. In other words, the bob moves up doing work against the gravitational force, losing its kinetic energy and gaining an equivalent potential energy.

If the maximum height attained by the doll is h (fig.), we have

½ mv² = mgh where m is the mas of the doll, v is the velocity of the truck and g is the acceleration due to gravity. Therefore, v = (2gh)^1/2 .

But h = ℓ – ℓ cosθ = ℓ(1– cosθ).

Substituting for h we have v = [2gℓ(1– cosθ)]^1/2

(2) The kinetic energy of a moving particle at time t is found to be directly proportional to t². The force acting on the particle is directly proportional to

(a) t²

(b) t^3/2

(c) t

(d) t^1/2

(e) t⁰

since the kinetic energy is directly proportional to t², we can write

½ mv² = C t² where m and v are the mass and velocity of the particle and C is the constant of proportionality.

Therefore, v = [√(2C/m)] t

The velocity is thus directly proportional to the time t which means that the particle is in uniformly accelerated motion. The acceleration is therefore constant. This means that the force (F = ma) is constant and hence independent of time. [Option (e)].

(3) A body of mass 5 kg at rest is acted on by two forces 3 N and 4 N which are at right angles to each other. The kinetic energy of the body at the end of 10 s is

(a) 250 J

(b) 350 J

(c) 500 J

(d) 750 J

(e) 100 J

The resultant force on the body is √(3² + 4²) = 5 N.

The acceleration of the body, a = F/m = 5 N/ 5 kg = 1 ms^–2

The velocity of the body at the end of 10 s is given by

v = v₀ + at with usual notations so that

v = 0 + 1×10 = 10 ms^–1

Therefore, the kinetic energy of the body, K = ½ mv² = ½ ×5 ×10² = 250 J.

(4) A steel ball bearing thrown vertically down bounces (from the floor) 5 m higher than its original level. With what speed was the ball bearing thrown down? Assume elastic collision at the floor and g = 10 ms^–.2

(a) 5 ms^–1

(b) 10 ms^–1

(c) 12 ms^–1

(d) 16 ms^–1

(e) 20 ms^–1

Since the collision at the floor is elastic, the ball bearing will return to the original level with the magnitude of its velocity unchanged, but the direction reversed. Therefore, the problem is equivalent to finding the velocity (v) with which the ball bearing is to be thrown vertically upwards to reach a height of 5 m.

Thus v = √(2gh) = √(2×10×5) = 10 ms^–1

[You can get the above relation by equating the initial kinetic energy to the gravitational potential energy at the maximum height h as ½ mv² = mgh or from the equation of motion, 0 = v² – 2gh].

(5) When a long spiral spring is stretched by 2 cm, its potential energy is U. If this spring is stretched by 3 cm, its potential energy will be

(a) 3U/2

(b) 9U/4

(c) 5U/2

(d) 3U

(e) 3U/2

We have U = ½ K×(0.02)² = 0.0002 K where K is the spring constant.

When the stretch is 3 cm, the potential energy is ½ K×(0.03)² = 0.00045 K = 9U/4

[It will be better to use the proportionality relation in situations like this:

U α 2²

X α 3²

Therefore, U/X = 4/9 so that X = 9U/4]

(6) The momentum of a car is increased by 3%. Because of this the kinetic energy of the car will be

(a) increased by 3%

(b) decreased by 3%

(c) increased by 6%

(d) decreased by 6%

(e) increased by 9%

The kinetic energy is directly proportional to the square of the momentum (K = p²/2m), Therefore, when the percentage increase in the kinetic energy will be twice the percentage increase in the momentum [Option (c)].

[The fractional change in kinetic energy is given by DK/K = 2Dp/p which is twice the fractional change in momentum. The percentage change in kinetic energy is therefore twice the percentage change in the momentum].

The following questions are for AP Physics C aspirants:

(7) A body suffers a displacement r under the action of a force F which varies inversely as the displacement r. The work done by the force is proportional to

(a) r⁰

(b) r

(c) r²

(d) 1/r

(e) ln r

We have W =∫F.dr

Since the force F and the displacement r are in the same direction

W =∫Fdr

But F = k/r where k is a constant.

Therefore, W = ∫Fdr = ∫kdr/r = k ln r

Thus the work done is proportional to the logarithm of displacement [Option (e)].

(8) A conservative force F = – 4(x î + y ĵ) newton moves a particle in the xy-plane. Initially the particle is at the origin. The work done by the force when the particle is given a displacement r = (2 î + 2 ĵ) metre is (î and ĵ are unit vectors along the x and y directions respectively)

(a) – 4 J

(b) – 8 J

(c) 8 J

(d) –16 J

(e) 16 J

You may imagine the particle to be moved first along the positive x-direction from the origin O to the point (2, 0) and then along the positive y-direction from the the point (2, 0) to the point (2,2) as shown in the figure. The value of x changes from zero to 2 during the motion along the x-direction and the value of y changes from zero to 2 during the motion along the y-direction.

Work done during the motion along the x-direction is

W₁ = ₀∫² F_x dx = ₀∫² (– 4x) dx = (– 4)×(2²/2) = – 8 J.

Work done during the motion along the y-direction is

W₂ = ₀∫² F_y dy = ₀∫² (– 4y) dy = (– 4)×(2²/2) = – 8 J.

Therefore, total work done = – 16 J.

In the next post we will discuss more questions in this section.

AP Physics B & C- Work, Energy & Power- Equations to be Remembered

You must remember the following points to make you strong in answering multiple choice questions involving work, energy and power:

(1) Work is done by a force if the point of application of the force is displaced. If the force vector F is constant and makes an angle θ with the displacement vector r, the work done (W) is given by

W = Fr cos θ

In vector notation this is given by

W = F.r

Thus work is a scalar quantity given by the scalar product of force and displacement.

If the force is not constant and is a function of the displacement (as for instance, in compressing or elongating a spring), the work done is given by W = ∫F.dr where dr is a small displacement for which the force F can be taken to be constant. The integration is to be carried out over the entire displacement. The work done by you in producing an elongation ‘x₁’ in a spring (which is not deformed initially) of force constant ‘k’ is given by

W = ₀∫^x1 F.dx = ₀∫^x1 kxdx = ½ kx₁²

[Note that the force with which you have to pull the spring to produce an extension ‘x’ is kx]

The work done in producing an extension ‘x’ in a spring is ½ kx².

If the elongation of a spring is to be increased from x₁ to x₂ the work required is ½ k(x₂² – x₁²).

If the force and the displacement are in the same direction and the magnitude of the force varies with displacement, the work done is given by the area under the force-displacement graph obtained by plotting the displacement on the x-axis and the force on the y-axis (fig.).

In the case of a spring the force-elongation (displacement) curve is a straight line as shown in the figure. The work done in producing an elongation ‘x’ in a spring is the area under the curve (which is the area of the triangle OAB) and is equal to ½ Fx = ½ (kx)x = ½ kx².

(2) A force is conservative if the work done by the force in moving an object depends only on the initial and final positions of the object and is independent of the path followed between these positions. The total work done by a conservative force on an object is zero when it moves round any closed path and returns to the initial position. Gravtational force and electrostatic force are examples of conservative forces.

If the work done by a force on an object moving between two positions depends on the path taken, the force is called non-conservative force. Friction and visous force are non-conservative forces.

(3) Kinetic energy (K) of a body of mass m moving with velocity v is given by

K = ½ mv²

Since momentum, p = mv, we have

K = p²/2m

[Remember this useful expression].

(4) Work-energy theorem states that the work done by a force acting on a body is equal to the change in the kinetic energy of the body. If W is the work done and K_i and K_f are the initial and final kinetic energies respectively, we have

W = K_f – K_i

(5) The gravitational potential energy of a body of mass m at a height h near the surface of the earth is mgh where g is the acceleration due to gravity at the place. Note that this energy is with respect to the reference level used for measuring the height and the value of h is negligible compared to the radius of the earth.

The change in the gravitational potential energy (∆U_g) of a mass m raised through a small height h can therefore be written as

∆U_g = mgh

You will find more details on gravitational potential energy in the posts dated 9^th, 12^th and 17^th May 2008 which you can access by clicking on the label ‘gravitation’ below this post.

(6) When a spring is compressed or elongated, the work done on the spring is stored as elastic potential energy in the spring. The elastic potential energy in the spring which is elongated or compressed through a distance ‘x’ is ½ kx².

(7) Law of conservation of energy states that energy can neither be created nor destroyed but can only be transferred from one form to another.

You should understand that the production of energy by annihilating mass in nuclear reactions does not violate the law of conservation of energy since mass itself is to be treated as a cocentrated form of energy in accordance with Einstein’s mass-energy relation, E = mc².

(8) Power is the time rate at which work is done or energy is transferred.

The average power P_av = W/t where W is the total work done in a time t.

Instantaneous power P = dW/dt where dW is the work done in a very small time dt at the instant t.

(9) The instantaneous power can be expressed as the scalar product of the force vector F and the velocity vector v as

P = F.v

This is easily obtained since dW = F.dr so that P = dW/dt = F.(dr/dt) = F.v

(10) Elastic collision is one in which momentum and kinetic energy are conserved. Inelastic collision is one in which kinetic energy is not conserved, but momentum is conserved.

You should note that momentum is conserved in elastic as well as inelastic collisions; but kinetic energy is conserved in the case of elastic collisions only.

In a completely inelastic collision the two colliding bodies move together after the collision.

If the mass m₁ moves with velocity v_1i in the positive x-direction and suffers a completely inelastic head on collision (collision in one dimension) with the mass m₂ at rest (fig.), the common velocity (v_f) with which the two masses will move is given by

m₁v_1i = (m₁ + m₂) v_f, on applying the law of conservation of momentum.

Therefore, v_f = m₁v_1i /(m₁ + m₂)

If the collision is elastic in the case of the above masses, we will have different

final velocities v_1f and v_2f for the two bodies after the collision, as given by the momentum conservationlaw:

m₁v_1i = m₁ v_1f + m₂ v_2f -------(i)

Since there are two unknowns (v_1f and v_2f) we require one more equation to solve for the unknowns and we have the kinetic energy equation,

½ m₁v_1i² =½ m₁ v_1f ²+ ½ m₂ v_2f ²

Or, m₁v_1i² = m₁ v_1f ²+ m₂ v_2f ²-------(ii)

Equations (i) and (ii) give

v_1f = (m₁– m₂) v_1i /(m₁ + m₂) and

v_2f = 2 m₁v_1i /(m₁ + m₂)

If the the two bodies are of the same mass, the colliding mass m₁ will come to rest and will hand over its velocity to the other mass m₂ which was initially at rest. If m₂>> m₁, the velocity of m₁ will get reversed and m₂ will continue to remain at rest.

If m₂ also has an initial velocity v_2i you can incorporate its initial momentum and kinetic energy in the above equations and solve for the final velocities of the bodies. You will then get

v_1f = [2 m₂v_2i + (m₁– m₂) v_1i] /(m₁ + m₂) and

v_2f = [2 m₁v_1i + (m₂– m₁) v_2i] /(m₁ + m₂)

Try to derive these equations as an exercise.

In the next post we will discuss questions in this section. Meanwhile go through some useful multiple choice questions (with solution) here at physicsplus.

Magnetic Force on Moving Charges- Two Questions for AP Physics B&C

"I believe in standardising automobiles, not human beings."

– Albert Einstein

Questions involving magnetic fields were discussed on this site on many occasions. You can access all of them by clicking on the label ‘magnetic field’ below this post. Today we will discuss two questions involving magnetic force on moving charges:

(1) A beam of positive ions of mass m, charge q and kinetic energy K proceeding along the positive x-direction emerges from the exit point O of an ion source. A plate P is placed perpendicular to the beam at a distance d from the point O (Fig.). A magnetic field is applied in the region between the point O and the plate P so that the ions are prevented from striking the plate P. Assuming that the ions are moving at non-relativistic speed, the minimum magnitude required for the magnetic field is

(a) [√(2 mK)] /qd

(b) [√(2 m²K)] /qd

(c) [√(2 K)] /qmd

(d) [√(2 mK² )] /qd

(e) [√(2 m)] /qKd

If the speed of the ion is v, we have

K = ½ mv² so that v =√(2K/m)

The beam will not strike the plate B if the radius of the path of the ions in the magnetic field is less than the distance d between the plate and the point O. If the magnetic field required is to be minimum, the field has to be applied perpendicular to the direction of the beam. This minimum field will make the radius R of the path equal to d. Therefore, we have

qvB = mv²/R = mv²/d so that

B = mv/qd

Substituting for v from the expression for kinetic energy we have

B = m√(2K/m)/qd = [√(2 mK)] /qd

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(2) A sensitive electronic balance (weighing machine) carries a magnet which produces a horizontal magnetic field B tesla in the central region between its poles N and S(fig.). The reading of the balance is R newton. A horizontal beam of negatively charged particles with a constant beam current is now passed continuously through the central region between the poles of the magnet. The direction of the beam is perpendicular to the magnetic field as shown. Then the reading of the electronic balance will

(a) be R itself

(b) be less than R

(c) be greater than R

(d) go on increasing from R

(e) go on decreasing from R

The beam of negative charges proceeding rightwards, as shown in the figure, is equivalent to a beam of positive charges proceeding leftwards. The magnetic force acting on the particles is downwards (as given by Fleming’s left hand rule). The charged particles exert an equal and opposite force (reaction) on the magnet. Therefore, there is an upward force on the magnet. The electronic balance therefore shows a reading less than R.

You will find some useful questions (with solution) in this section at physicsplus