A 0.36 kg particle has a speed of 6.1 m/s at point A and kinetic energy of 11.3 J at point B. What is the kinetic energy at A. What is the speed at B. What is the work done in moving from A to B.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The particle has a mass of 0.36 kg. At point A, the speed of the particle is 6.1 m/s. Its kinetic energy at point B is 11.3 J.

The kinetic energy of a particle with mass m and moving at a speed v is given by K.E. = (1/2)*m*v^2

As the speed of the particle is 6.1 m/s at point A, its kinetic energy is (1/2)*0.36*(6.1)^2 = 6.6978 J

The kinetic energy of the particle at B is given as 11.3 J. If the speed of the particle at point B is v, (1/2)*0.36*v^2 = 11.3

=> v^2 = 2*11.3/0.36 = 565/9 m/s

=> v = 7.92 m/s.

The work done by the particle in moving from A to B is the difference in the energy of the particle at the two points. This is equal to 11.3 - 6.6978 = 4.6022. The work done by the particle in moving from A to B is 4.6022 J.

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