Question

The rate of decrease of kinetic energy is 9.6j/s. Find magnitude of force acting on particle when it�s speed in 3 m/s?

• A. 3.2N
• B. 4.8N
• C. 2.4N
• D. 5.6N

Answer 1

Answer: (A)

3.2N

Answer 2

Given, and $\frac{dK}{dt}=9.6$ j / sec $\quad\quad$v = 3 m/sec kinetic energy is given by, $K=\frac{1}{2}$ mv Rate of change of kinetic energy is, $\frac{dK}{dt}=\frac{1}{2}m\times2v\times\frac{dv}{dt}$ $\frac{dK}{dt}=v\times\frac{mdv}{dt}$ $\Rightarrow \frac{dK}{dt}=v\times$ Force Force$=\left(\frac{dK}{dt}\right) \frac{1}{v}=\frac{9.6}{3}=3.2$ Force = 3.2 N