# Concept of work and Work energy Theorem

• Work is the dot product of force and displacement.
• W = F S cos θ.
• Work is a scalar quantity.
• If 0° ≤ θ < 90° work is positive.
• If 90° < θ ≤ 180° work is Negative.
• If θ = 90° work is zero.
• When a body moves in a circular path work done by centripetal force is zero.
• When lift moves up with acceleration ‘a’ work done by tension is W = m (g + a) h.
• When lift moves down with acceleration ‘a’ work done by tension is W = m (g − a) h.
• Body moving on a horizontal surface of coefficient of friction "μ" work = μ mgs.
• Work done in lifting a body of density ‘d1’ lies in a liquid of density ‘d2’ work in lift = mgh (1 – d2/d1).
• A bucket full of water of mass ‘M’ is lifted with a slope of mass m work done = Mgl + \tt \frac{mgl}{2}.
• Work done in lifting a pendulum bob pulled aside through an angle ‘θ’ with vertical by a horizontal force F. W = − mgl (1 – cos θ).
• Work done in lifting a uniform ‘rod’ pulled aside through an angle ‘θ’ with vertical \tt W = -mg \frac{L}{2} (1 – cos θ).
• Work done by gravity in lifting a rod to held at angle ‘θ’ with surface \tt W = \frac{-mg|}{2} \sin \theta
• Work done in lifting a chain of mass ‘m’ length 'l' such that bottom touches the ground = \tt mg\frac{|}{2}
• A chain of mass ‘m’ and length ‘l’ on the table and \tt \frac{1}{n^{th}} part is hanging from edge, work done is in pulling the chain on to table \tt W = \frac{mg|}{2n^{2}}
• Work done by a variable force ∫dw = ∫F·ds.
• Area under force displacement graph gives work done.
• CGS unit of work is erg.
• 1 erg = 1 dyne × 1 cm.
• S.I unit of work is Joule.
• 1 Joule = 1 N × 1 cm.
• 1 Joule = 107 erg.
• Work done by variable force \tt W=\lim_{x \rightarrow 0} \int_{xi}^{xf} Fdx.
• Work done in stretching a spring \tt W = \frac{1}{2} kx^{2}. {K = spring constant}.
• According to work energy theorem work done by a net force acting a body is equal to the change produced in kinetic energy.
• Work energy theorem \tt W = \frac{1}{2} mv^{2} - \frac{1}{2} mu^{2}.
• When only internal force is acting Winternal = ΔKE.
• Both internal and external are acting Wint + Wext = ΔKE.
• When different types of forces are acting on a body  Wext + Wint + Wpseudo + Wother = ΔKE. = \tt \frac{1}{2} mv^{2} - \frac{1}{2} mu^{2}.
• Work energy theorem is a scalar form of second law.
• Work energy theorem holds in all inertial frames.
• In Non-inertial frame pseudo force has to be included.
• When force and displacement are in the same direction the kinetic energy of the body increases.
• When force and displacement are in the opposite direction the kinetic energy of the body decreases.
• Work done by centripetal force is zero.
• 1ev = 1.6 × 10−19 Joules.
• Kinetic energy \tt K.E=\frac{p^{2}}{2m} (P = momentum)
• Kinetic energy \tt K.E=\frac{1}{2} mv^{2}
• Momentum \tt P = \sqrt{2 m KE}
• Percentage change in momentum = \tt \frac{\Delta P}{P} \times 100 = \frac{1}{2} \frac{\Delta KE}{KE} \times 100
• A body of mass ‘m’ falls from height ‘h’ on to a spring of spring constant mg (h + x) = \tt \frac{1}{2} Kx^{2}.
• When released just on spring \tt mgx = \frac{1}{2} Kx^{2}.
• When a body is throws up with velocity ‘u’ from height ‘h’. \tt \frac{1}{2} mu^{2} + mgh = \frac{1}{2} mv^{2} (u = final velocity).

### View the Topic in this video From 01:00 To 27:23

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1. If under a constant force F the object displaced through a distance s, then work done by the force W = F · s = Fs cosθ

2. Work done by the force of gravity on a particle of mass m is given by W = mgh

3. Work done in compressing or stretching a spring is given by \tt W= \frac{1}{2}\ kx^{2}

4. Work done by the couple for an angular displacement θ is given by W = τ · θ

5. Kinetic energy of an object is given by, K=\frac{1}{2}\ mv^{2}=\frac{p^{2}}{2m}

6. Work done by a force is displacing a body is equal to change in its kinetic energy.
W=\int_{v_{1}}^{v_{2}}F\cdot ds=\frac{1}{2}\ mv_2^2-\frac{1}{2}\ mv_1^2=K_{f}-K_{i}=\Delta KE
where, Ki = initial kinetic energy and kf = final kinetic energy.

7. Stopping distance (x) = \tt \frac{Kinetic\ energy(E)}{Stopping\ force(F)}\Rightarrow x=\frac{mv^{2}}{2F}

8. Stopping time t=\frac{mv}{F}

9. Comparison of stopping distance and time for two vehicles: The ratio of their stopping distances \tt \frac{x_{1}}{x_{2}}=\frac{E_{1}}{E_{2}}=\frac{m_{1}v_1^2}{m_{2}v_2^2}
The ratio of their stopping time \tt \frac{t_{1}}{t_{2}}=\frac{P_{1}}{P_{2}}=\frac{m_{1}v_1}{m_{2}v_2}