There are many forms of potential energy these

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There are many forms of potential energy. These include: (i)Gravitational (ii)Elastic (iii)Chemical GRAVITATIONAL POTENTIAL ENERGY This is the energy possessed by an object due to its position in the gravitational field. It is given by the equation; P.E = mgh KINETIC ENERGY This is the energy possessed by an object by reason of its motion K.E is given by the formula; K.E = mv 2 REST ENERGY This is the energy possessed by an object by reason of its mass alone It is given by the formula; E = mc 2 Where E = rest energy m= mass of object C= speed of light EXAMPLE
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Prepared By: ChilesheMwenyaMporokoso Page 58 Consider the pendulum shown below. If it is released at rest from a point A, How fast is the ball moving (a) At B (b)At C H = 1.5Cos40 o = 1.15 h = 1.5 1.15 = 0.35 P.E at A = m g h = m x 9.81 x 0.35 = 3.4335m In moving from A to B P.E at A is converted to K.E at B
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Prepared By: ChilesheMwenyaMporokoso Page 59 P.E (at A) = K.E (at B) 3.4335m = mv 2 V = √6.867 V = 2.62m/s At C V= 0m/s because at B it momentarily stops and starts moving back EXERCISE An object is released from rest at the top of a tower of height 47m and falls to the ground,Calculate; (i)The time it takes to fall to the ground (ii)Its speed just before it hits the ground. WORK DONE BY A CONSTANT TORQUE A torque or a moment of force is the product of the force and the perpendicular distance from the axis of rotation to the force. If F is the tangential force acting on a point A, W = F x s ………………………………………….(i) But Torque T = Fr……………………………(ii) S = r θ …………………………………………………..(iii) Sub (iii) into (i) W = F x r θ . Fr = T
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Prepared By: ChilesheMwenyaMporokoso Page 60 W = T θ EXAMPLE An electric motor has to lift an object hanging a mass of 50kg by means of a rope wound round a drum having a diameter of 1.2m. Calculate; (i)The torque to be exerted (ii)The work done when the drum makes 20 revolutions Solution D = 1.2 , r = ଵ.ଶ = 0.6m (i)T = F x r, F = mg 50 x 9.81= 490.5N T = 490.5 x 0.6 = 294.3Nm (ii)W = T x θ = 294.3 x 20 x 2 π = 36 982.83J POWER REQUIRED FOR ROTATION Power = ௪௢௥௞ௗ௢௡௘ ௧௜௠௘ = ்ఏ T x but = ω Power = T x ω If ω is in rads/s then use the above formula If ω is in rev/min then use P = ଶగே் ଺଴
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Prepared By: ChilesheMwenyaMporokoso Page 61 EXAMPLE A rotor transmits a torque of 1400Nm at a speed of 660 r.p.m, Calculate; (i)The angular velocity of the shaft (ii)The power transmitted Solution ߱ = 660rev/min = ଺଺଴௥௘௩ ଺଴௦ = 11rev/s = 11 x 2 π = 69.12 rads/s (ii) P = ଶగே் ଺଴ = ଶగ௫ ଺଺଴ ௫ ଵସ଴଴ ଺଴ = 96 761.054W
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Prepared By: ChilesheMwenyaMporokoso Page 62 UNIT 1.9: STRAIN AND STRESS STRESS:(or Pressure) is define as force per unit cross section area Stress( σ ) = ி௢௥௖௘ ஼௥௢௦௦ି௦௘௖௧௜௢௡஺௥௘௔ = ி The units of measurement for stress are N/m 2 or Pa. Example A tier bar has a cross sectional area of 125m 2 and is subjected to a pull of 10KN. Calculate the stress in meganewtons per square metre. Solution A = 125m 2 , 1000mm 2 = 1m 2 125m 2 x ଵ଴଴଴଴଴଴௠ = 125 x 10 6 m 2 σ = ி = ଵ଴଴଴ ଵଶହ ௫ ଵ଴ షల = 80 x 10 6 N/m 2 = 80MN STRAIN: This is the ratio of the change in dimension to the original dimension.
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