6At the time tl \"aanobject ,.XNew Tab t-Responses/submit?dep 110% | … 65.04 kJ

Question

6At the time tl "aanobject ,.XNew Tab t-Responses/submit?dep 110% | … 65.04 kJ eBook Notes Ask Your At the time ti position 0, an object with a mass m 5.01 kg passes th (8.ooi + 4.00j) m. Assume that the force acting on the object during the time interval &t-t; is constant and determine the rough the origin with a velocity Vi (5.61i -2.01j) m/s. At a time t2 2.00 s, the object is at following. (a) the kinetic energy of the object at the time t (b) the force acting on the object during the time interval t2-t1 (Express your answer in vector form.) (c) the work done on the object by the force during the time interval -Q- (d) the kinetic energy of the object at the time t2 (e) the speed of the object at the time t m/s eBook Tutorial

Explanation / Answer

a)speed of the object = (vx^2 + vy^2)^0.5

=(5.61^2+2.01^2)^0.5

=5.959 m/s

so kinetic energy = 0.5mv^2

=0.5*5.01*5.959^2

=88.95 J

b)let the acceleration in the x direction be x and that in the y direction be y.so, for thx direction,

S = ut + 0.5at^2

or 8 = 5.61*2 + 0.5*x*2^2

or x = -1.61 m/s^2

for the y direction,

4 = -2.01*2 + 0.5*y*2^2

or y=4.01 m/s^2

so Force = ma

F = -8.066 i + 20.09 j (m/s^2)

c)Work done = F.dS

=(-8.066*8 + 20.09*4)

=15.83 J

d)let the final kinetic energy of the object be KE.so,

KE = initial energy + work done

=88.95 + 15.83

=104.78 J

e)0.5mv^2=KE

or 0.5*5.01*v^2 = 104.78

or v=6.46 m/s

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