question_answer

A particle of mass 50 g moves on a straight line. The variation of speed with time is shown in figure. Find the force acting on the particle t = 2, 4 and 6 seconds.

Answer:

Given \[m=50\,\,g=\frac{40}{1000}kg=\frac{5}{100}kg,\,\,F=?\] (a) Slope of at straight line \[OA=\frac{15}{3}=5\] Acceleration at 2 seconds \[=5m/{{s}^{2}}\] Force at 2seconds\[=50\times {{10}^{-3}}\times 5\] \[=250\times {{10}^{-3}}N=0.25\,\,N\]along the motion (b) At 4 seconds slope is zero and hence acceleration is zero. \[\therefore \] force at 4 seconds = zero (c) Slope of BC straight line\[=\frac{0-15}{3}=-5\]which gives the acceleration any point between \[BC=-5m/{{s}^{2}}\] \[\therefore \]force at 6 seconds or acceleration \[=-50\times {{10}^{-3}}\times 5\] \[=-250\times {{10}^{-3}}N=0.25N\]opposite to the motion