CBSE Class 9 CBSE Class 9 Science
(a) State the law of conservation of energy. (b) What is the work done to increase the velocity of a car from 36 km h^-1 to 72 km hr^-1 if the mass of the car is 1500 kg ? Does the work done by the force have a negative or a positive magnitude ? © Where does an oscillating pendulum have maximum PE and KE ?
(a) It states that energy can neither be created nor be destroyed. (b) Given m = 1500 kg, u = 36 km h^-1 = 10 m s^-1, v = 72 km h^-1 = 20 m s^-1, W = ? Work done = Change in kinetic energy =1/2 m (v^2-u^2)=1/2 x 1500 x (20^2 -10^2) = 225000 J The work done is positive. © It has maximum potential energy at its extreme position and maximum kinetic energy at the mean position.
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Concept:
W = Kf - Ki \(W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} = {\bf{\Delta }}K\,\) Where v = final velocity, u = initial velocity and m = mass of the body Calculation: Given - mass (m) = 1500 kg, initial velocity (u) = 36 km/h = 10 m/s and final velocity (v) = 72 km/h = 20 m/s
\(\Rightarrow W = \frac{1}{2}m({v^2} - {u^2} )\,\) \(\Rightarrow W = \frac{1}{2}\times 1500\times ({400} - {100} )=2.25\times10^5\, J\,\) India’s #1 Learning Platform Start Complete Exam Preparation
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Given, Mass of the car, m = 1500 kg Initial velocity of the car, u = 30 km/h = $\frac{30\times 1000m}{3600s}=\frac{25}{3}m/s$ [converted km/h to m/s] Final velocity of the car, v = 60 km/h = $\frac{60\times 1000m}{3600s}=\frac{50}{3}m/s$ [converted km/h to m/s] To find = Work done (W) Solution: According to the Work-Energy theorem or the relation between Kinetic energy and Work done - the work done on an object is the change in its kinetic energy. So, Work done on the car = Change in the kinetic energy (K.E) of the car = $Final\ K.E-Initial\ K.E$ $Work\ done, \ W =\frac{1}{2}m{v}^{2}-\frac{1}{2}m{u}^{2}$ $[\because K.E=\frac{1}{m}{v}^{2}, \ where, \ mass\ of\ the\ body=m,\ and\ the\ velocity\ with\ which\ the\ body\ is\ travelling=v]$ $W=\frac{1}{2}m[{v}^{2}-{u}^{2}]$ $[taking\ out\ common]$ Now, substituting the values- $W=\frac{1}{2}\times 1500[(\frac{50}{3}{)}^{2}-(\frac{25}{3}{)}^{2}]$ $W=\frac{1}{2}\times 1500[(\frac{50}{3}+\frac{25}{3})(\frac{50}{3}-\frac{25}{3})]$ $[\because ({a}^{2}-{b}^{2})=(a+b)(a-b)]$ $W=\frac{1}{2}\times 1500\times \frac{75}{3}\times \frac{25}{3}$ $W=156250J$ Hence, the work to be done to increase the velocity of a car from 30km/h to 60km/h is 156250 joule, if the mass of the car is 1500 kg. |
What work has to be done to increase the velocity of a car from 36km h to72km hif the mass of the car is 1500kg?
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