A particle of mass m is moving along a trajectory given by `x=x_0+a cosomega_1t` `y=y_0+bsinomega_2t` The torque, acting on the particle about the origin, at `t=0` is:
The coordinates of a particle of mass 'm' as function of time are given by x=x_(0)+a_(1) cos(omegat) and y=y_(0)+a_(2)sin(omega_(2)t) . The torque on particle about origin at time t=0 is :
The position (x) of a particle of mass 2 kg moving along x-axis at time t is given by x=(2t^(3)) metre. Find the work done by force acting on it in time interval t=0 to t=2 is :-
The potential energy of a particle of mass 5 kg moving in xy-plane is given as U=(7x + 24y) joule, x and y being in metre. Initially at t=0 , the particle is at the origin (0,0) moving with velovity of (8.6hati+23.2hatj) ms^(1) , Then
The position (x) of a particle of mass 1 kg moving along X-axis at time t is given by (x=1/2 t^(2)) metre. Find the work done by force acting on it in time interval from t=0 to t=3 s .
The position (x) of a particle of mass 1 kg moving along x-axis at time t is given by (x=(1)/(2)t^(2)) metre. Find the work done by force acting on it in time interval from t=0 to t=3 s.
The coordinates of a moving particle at any time 't' are given by x = alpha t and y = beta t . The speed of the particle at time 't' is given by
The displacement 'x' of a particle moving along a straight line at time t is given by x=a_(0)+a_(1)t+a_(2)t^(2) . The acceleration of the particle is :-
A time varying force, F=2t is acting on a particle of mass 2kg moving along x-axis. velocity of the particle is 4m//s along negative x-axis at time t=0 . Find the velocity of the particle at the end of 4s.
