If a #2 kg# object is constantly accelerated from #0m/s# to #16 m/s# over 6 s, how much power musy be applied at #t=3 #?

3 Answers
Jan 20, 2018

Instantaneous power #P=42.666W#atts.

Explanation:

When a body of mass#'m'# is initially at rest, is moved by the application of a constant force, its velocity changes to #'v_o'# in time #'t_o'#,then Instantaneous power at any time #'t'# generated is given as #P=m(v_o/t_o)^2t#
Substituting the given values we get,
#P=2kg[(16m//s)/(6s)]^2xx3s=2kg[64m^2//9s^3]xx3s=(128kgm^2)/(3s^3)=42.666W#
#:.P#ower#=42.6W#atts.

Jan 20, 2018

THe power is #=42.67W#

Explanation:

To calculate the acceleration, apply the equation of motion

#v=u+at#

The initial velocity is #u=0ms^-1#

The final velocity is #v=16ms^-1#

The time is #t=6s#

The acceleration is

#a=(v-u)/t=(16-0)/6=8/3ms^-2#

The velocity at #t=3s# is

#v(3)=u+at#

#v(3)=0+8/3*3=8ms^-1#

The mass is #m=2kg#

According to Newton's Second Law

#F=ma#

The force is #F=2*8/3=16/3N#

The power is

#P=Fv=16/3*8=42.67W#

Jan 20, 2018

43 W at t=3s or
#22# W average power in the first 3 s.

Explanation:

#barP= "Average Power" = "Work"/"Time duration" =W/(Deltat) #.

#because W= FDeltax # if F is a constant.

# bar P = W/(Deltat) = F(Deltax)/(Deltat) = F barv #

The instantaneous power at a given time is then

#P = lim_(Deltat rarr0) F(Deltax)/(Deltat)=F v#

Since the object is accelerating

#F =F_"net"= ma = m (v-v_0)/t = 2((16-0)/6) = 5.33N #
#v_0=0 rArr#
#v= v_0+ at =at#
and
#P = Fv = (ma)at = ma^2t #

at t = 3s

#P = ma^2t= 2kg*((16m/s-0)/(6s))^2*3s =42.7 (kgm^2/s^2)/s ~~ 43 "Watts"#

Bear in mind that this is instantaneous power. If the question meant for average power delivered from t=0 to t=3 s, then

#barP ~~½ (43) ~~ 22 W#

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