Basic Mechanics Tutorials- At a glance
Work, Force and Energy - with Examples, Problems, Solutions- Try out the MCQ Quiz questions at the end
Target Audience: High School Students, College Freshmen and Sophomores, students preparing for the International Baccalaureate (IB), AP Physics B, AP Physics C, A Level, Singapore/GCE A-Level;
The terms work, force and energy sound very familiar. We use them frequently in our everyday life. We associate work with a lot of different activities like a mother cooking food, a student studying, children playing cricket, etc. A force is a push or pull on an object. Energy is the capacity to do work. This shade of meanings is close to that of their meanings in physics, but in contrast their meanings in physics are very definite and precise.
Work is defined as energy transfer from one object to another by means of a force. Energy transferred to the object is positive work and energy transferred from the object is negative work.
The work done on a particle by a constant force during displacement ‘d’ is
W = F.d = Fdcosθ
in which θ is the constant angle between the directions of F and d. Only the component of F that is along the displacement can do work on the object. When two or more forces act on an object, their net work is the sum of the individual works done by the forces, which is also equal to the work that would be done on the object by the net force of those forces.
The work-energy theorem states “For a particle, a change ∆K in the kinetic energy equals the net work W done on the particle”.
We know that, W = ∫F.dr = ∫Fcosθ dr …(1)
where the integration is performed along the path of the particle.
Also, dK/dt = d(½ mv2)/dt = mv dv/dt = F v
where F is the resultant force in the direction of the velocity
Therefore, dK/dt = F.v = F.dr/dt
which gives dK = F.dr …(2)
Putting (2) in (1);
W = ∫dK = K2 – K1 = ∆K
Thus, the work done by the resultant force on a particle is equal to the change in its kinetic energy.
The force from a spring is, F = -kx (Hooke’s Law)
The work done on an object by the spring force attached to the spring’s free end, when the object is moved from an initial position xi to a final position xf is
W = ∫ -kx dx
where dx varies from xi to xf.
Therefore, W = - ½ kx2 (from xi to xf) = ½ kxi2 – ½ kxf2
Potential energy is a kind of energy which depends on the configuration of the system. The earth attracts every object by a force ‘mg’ towards it. Gravitational potential energy is defined corresponding to the work done by this force. Thus, if an object rises above the surface of the earth, the potential energy of the ‘earth + object’ system rises by ‘mgh’. Similarly, if the object descends by ‘h’, the potential energy decreases by ‘h’. Here the frame of reference is attached to the earth as the acceleration of the earth compared to the object is very negligible, so it will very nearly be an inertial frame. The work done by the block on the earth in this frame is thus zero.
Suppose a particle moves along the curve AB and that AB makes an angle θ with the vertical. The force of gravity = mg acts on the object.
The work done this force = ∫F.dr = mg∫dr (as the force is constant in magnitude)
Now, as can be seen from the figure, ∫dr = ABcosθ
Therefore, W = mg(AB)cosθ = -mgh
If the particle goes from A to B along any other curve also, the work done by gravity is again mgh. If the particle starts from A and comes back to A after some time, the work done by gravity is zero as the change in height is zero. This kind of forces are called conservative forces.
If the work done by a force during a round trip of a system is always zero, the force is said to be conservative. Otherwise, it is non-conservative. It implies that the work done by a conservative force depends only on the initial and final states and not on the path taken. Basically, conservative forces convert energy into forms which can be stored in the system itself e.g; they convert kinetic energy into potential energy etc. in which case the kinetic energy can be gained back by losing potential energy (the system has to be defined carefully; here the earth is included in the system). The energy is not lost to the surroundings whereas non-conservative forces like friction convert energy into forms like heat energy and sound energy which are lost from the system and go into the surroundings.
The total energy E of a system (the sum of its mechanical energy (kinetic and potential energy) and its internal energies, including thermal energy) can change only by amounts of energy that are transferred to or from the system. This experimental fact is known as the law of conservation of energy. If work W is done on the system, then
W = ∆E = ∆Emec + ∆Eth + ∆EintIf the system is isolated (W = 0), this gives
∆Emec + ∆Eth + ∆Eint = 0This law is based on the fundamental fact that energy can neither be created nor destroyed. It can only be converted from one form to another. If no work is done on a system i.e. no energy is transferred to or from the system, then its total energy has to remain constant, it cannot change.
A very common confusion here is that whenever we apply the conservation of energy to a ‘given system + earth’ system that involves changes in potential energy, why do we not include the changes in the energy of the earth because there is an equal and opposite force acting on the earth? The concept to keep in mind here is that we take the frame of reference to be attached to the earth as the acceleration of the earth is infinitesimally small so this frame is essentially inertial. Hence the work done on the earth in this frame is zero.
Work, Calculation of work done, work-energy theorem,work done by a spring force, gravitational potential energy, work done by the force of gravity, conservative and non-conservative forces, conservation of energy, kinetic energy
A quick look at some of the questions solved in this tutorial :
Q: A single force acts on a 2.0 kg particle whose position is given by x = t + 5t2 + 2t3, with x in meters and t in seconds. Find the work done on the particle by the force from t = 0 to t = 3 s.
Q: A breadbox is made to move along an x axis from x = 0.2 m to x = 1.50 m by a force with a magnitude given by F = exp(-2x), with x in meters and F in newtons. How much work is done on the breadbox by the force?
Q: A conservative force F(x) acts on a 1 kg particle that moves along x axis. The potential energy U(x) associated with F(x) is graphed in the given figure. When the particle is at x = 2 m, its velocity is -2 m/s. What are the (a) magnitude and (b) direction of F(x) at this position? Between what positions on the (c) left and right does the particle move? (d) What is the particle’s speed at x = 8 m?
Q: A 2 kg object is acted on by a conservative force given by F = 4x - 3x2, with F in newtons and x in meters. Take the potential energy associated with the force to be zero when the object is at x = 0. (a) What is the potential energy of the system associated with the force when the object is at x = 1.7 m? (b) If the object has a velocity of 5 m/s in the negative direction of the x axis when it is at x = 6 m, what is its speed when it passes through the origin?
Q: A block of mass 5 kg is kept over another block of mass 10 kg and the system rests on a horizontal surface. A constant horizontal force F acting on the lower block produces an acceleration of 2 ms-2 in the system, the two blocks always move together. (a) Find the magnitude of force F. (b) Find the coefficient of kinetic friction between the bigger block and the horizontal surface. (c) Find the frictional force acting on the smaller block. (d) Find the work done by the force of friction on the smaller block by the bigger block during a displacement of 3 m of the system. Take g = 10 ms-2.
Q: In figure, a cord runs around two massless, frictionless pulleys. A canister with mass m = 15 kg hangs from one pulley, and you exert a force on the free end of the cord. (a)What must be the magnitude of F to lift the canister at a constant speed? (b) To lift the canister by 5 cm, how far must you pull the free end of the cord? During that lift, what is the work done on the canister by (c) your force (via the cord) and (d) the gravitational force?
Q: Two blocks, of masses 2.5 kg and 5 kg, are connected to a spring of spring constant k = 150 N/m that has one end fixed. The horizontal surface and the pulley are frictionless, and the pulley has negligible mass. The blocks are released from rest with the spring relaxed. (a) What is the combined kinetic energy of the two blocks when the hanging block has fallen 0.5 m? (b) What is the kinetic energy of the hanging block when it has fallen that 0.5 m? (c) What maximum distance does the hanging block fall before momentarily stopping?
Q: A block slides down an incline. As it moves from point A to point B, which are 6 m apart, a force of 5 N directed down the incline acts on the block. The magnitude of the frictional force acting on the block is 10 N. If the kinetic energy of the block increases by 45 J between A and B, how much work is done on the block by the gravitational force as the block moves from A to B?
Q: A 10 kg stone is at rest on a spring. The spring is compressed 10 cm by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional 20 cm and released. What is the elastic potential energy of the compressed spring just before that release? (c) What is the change in the gravitational potential energy of the stone–Earth system when the stone moves from the release point to its maximum height? (d) What is that maximum height, measured from the release point?
Q: A massless rigid rod of length L has a ball of mass m attached to one end. The other end is pivoted in such a way that the ball will move in a vertical circle. First, assume that there is no friction at the pivot. The system is launched downward from the horizontal position A with initial speed v0. The ball just barely reaches point D and then stops. (a) Derive an expression for v0 in terms of L, m, and g. (b) What is the tension in the rod when the ball passes through B? (c) A little grit is placed on the pivot to increase the friction there. Then the ball just barely reaches C when launched from A with the same speed as before. What is the decrease in the mechanical energy during this motion? (d) What is the decrease in the mechanical energy by the time the ball finally comes to rest at B after several oscillations?
Q: A spring with k = 200 N/m is at the top of a frictionless incline of angle 30°. The lower end of the incline is distance D = 1 m from the end of the spring, which is at its relaxed length. A 2 kg canister is pushed against the spring until the spring is compressed 0.2 m and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?
Q: A 500 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 300 N/m. The block becomes attached to the spring and compresses the spring 15 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible) (d) If the speed at impact is doubled, what is the maximum compression of the spring?
Q: A block of mass 0.5 kg is kept on a vertical spring of spring constant 80 N/m fixed from below. The spring is compressed to have a length which is 15 cm shorter than its natural length and the system is released from this position. How high does the block rise? Take g = 10 ms-2.
Q: The system shown in figure is released from rest and the block of mass 2 kg is found to have a speed of 0.5 m/s after it has descended through a distance of h = 1 m. Find the coefficient of kinetic friction between the block and the table. Take g = 10 ms-2.
Q: A simple pendulum of length 1 m having a bob of mass 0.5 kg is deflected from its rest position by an angle θ and released. The string hits a peg which is fixed at a distance x = 60 cm below the point of suspension and the bob starts going in a circle centred at the peg. (a) Assuming that initially the bob has a height less than the peg, what is the minimum height reached by the bob? (b) If the pendulum is released with θ = 90o and the peg was located 50 cm below the point of suspension, find the maximum height reached by the bob above its lowest position before the string becomes slack.
Q: A chain of length 80 cm and mass 2 kg lies on the surface of a smooth sphere of radius 1 m with one end tied to the top of the sphere. (a) Find the gravitational potential energy of the chain with reference level at the centre of the sphere. (b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain when it has slid through an angle 30o. Take g = 10 ms-2.
Q: A particle of mass 1 kg is kept on a fixed, smooth sphere of radius 0.5 m at a position, where the radius through the particle makes an angle of 30o with the vertical. The particle is released from this position. (a) What is the force exerted by the sphere on the particle just after the release? (b) Find the distance travelled by the particle before it leaves contact with the sphere. Take g = 10 ms-2.
Q: One end of a spring of natural length ‘l’ and spring constant ‘k’ is fixed at the ground and the other is fitted with a smooth ring of mass ‘m’ which is allowed to slide on a horizontal rod fixed at a height ‘l’. Initially the spring makes an angle 30o with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical. Take g = 10
Q: A uniform chain of length l and mass m overhangs a horizontal table with its three-fourth part on the table. The friction coefficient between the chain and the table is μ. Find the work done by the friction during the period the chain slips off the table.
Q: A smooth sphere of radius R is made to translate in a straight line with a constant acceleration ‘a’. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides.
Q: A particle of mass 2 kg is suspended by a string of length 1m. The particle is given a horizontal velocity, v0. The string becomes slack at some angle and the particle proceeds on a parabola. Find the value of v0 if the particle passes through the point of suspension. Take g = 10 ms-2.
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