Certain force acting on a 20kg mass changes its velocity from 5m/s to 2m/s. Calculate the work done by the force.
Answers
Answered by
19
Work done= Final Kinetic energy - Initial Kinetic Energy
= 1/2 mv^2 - 1/2 mu^2 = 1/2 *m ( v^2 -u^2)
= 1/2 *m (2^2 - 5^2)
= 1/2 *20 (4 - 25)
= 10 * -21
= -210 J
But as it is negative sign here it means that the work done was negative work.
= 1/2 mv^2 - 1/2 mu^2 = 1/2 *m ( v^2 -u^2)
= 1/2 *m (2^2 - 5^2)
= 1/2 *20 (4 - 25)
= 10 * -21
= -210 J
But as it is negative sign here it means that the work done was negative work.
Answered by
11
m = 20 kg
u = 5m/s
v = 2m/s
Work done = Change in Kinetic Energy
= Final K.E - Initial K.E
= (1/2)mv^2 - (1/2)mu^2
= (1/2)m (v^2-u^2)
= (1/2)×20×(4-25)
= 10 × (-21)
= -210 J
Hence the work done by The force is -210 J.
negative sign indicates negative work done, that is the body in not moving in the direction of force applied.
u = 5m/s
v = 2m/s
Work done = Change in Kinetic Energy
= Final K.E - Initial K.E
= (1/2)mv^2 - (1/2)mu^2
= (1/2)m (v^2-u^2)
= (1/2)×20×(4-25)
= 10 × (-21)
= -210 J
Hence the work done by The force is -210 J.
negative sign indicates negative work done, that is the body in not moving in the direction of force applied.
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