Question

A 5.97-kg object passes through the origin at time t = 0 such thatits x component of velocity is 4.70 m/s and its y component ofvelocity is -3.12 m/s.

(a) What is the kinetic energy of the object at this time?

(b) At a later time t = 2.00 s, the particle is located at x = 8.50m and y = 5.00 m. What constant force acted on the object duringthis time interval? (Magnitude ANDdirection)

A) we know that K.E is given by = (1/2)mv2
now, velocity = (vx2 + vy2)1/2
= 5.64
so, K.E = 1/2 x 5.97 x 5.642
= 94.99 J

B) now, at time t = 2 s
particle is located at x = 8.5 m and y = 5 m so, we have
acceleration in x direction = 2(s - ut ) /t2
= 2 x ( 8.5 - 4.7 x 2)/ 4
= -0.45 m/s2
now, acceleration in y direction = 2( 5 + 3.12 x 2 )/4
= 5.62 m/s2
so total acceleration =(ax2 + ay2)1/2
= 5.63 m/s2
so, now force = ma
= 5.97 x 5.63 N
= 33.65 N


What do I do to get the direction of this force?Only part I needed and was left out of solution when someoneprevious asked :P Degrees from x-axis.