Question icon
Grade 12Mechanics

under the action of a force, 2 kg body moves such that its position x as a function of time is given by x = (t)^(2/3) where x is in metres and t is in seconds. the work done by the force in the first two seconds is ?

Profile image of Nandit Sai
9 Years agoGrade 12
Answers icon

1 Answer

Profile image of Vikas TU
9 Years ago
velocity = dx/dt = (2/3)*t^-1/3
acceleration = (-2/9)*t^-4/3
Put t=2 in both v and accln.
Work =  Force x Velocity
F=ma and velocity has been calculated already.