If a #12 kg# object is constantly accelerated from #0m/s# to #4 m/s# over 7 s, how much power musy be applied at #t=1 #?

2 Answers
Apr 6, 2016

Answer:

#P(1) = 16/49J/s= 16/49 "Watts"#

Explanation:

Given: Object mass, #m= 12kg#, #v_i=0, v_f=4m/s#

Required: Power applied or use up at #t=1#

Solution Strategy:
a) Kinematic equation to determine the acceleration: #v_f= v_i +at#
b) Find the kinetic energy at any timer #1/2v^2(t)#
c) Calculate Power using: #P(t)= (dW)/dt|_(t=1)=P(1)#

#color(brown)("a)")#
#v(t) = at; v(7)=v_f=4=7a#
#a=4/7 m/s^2# thus
#v(t) = 4/7t #

#color(brown)("b)")# Kinetic energy = Work
#W=1/mv^2(t)= 1/2*(4/7t)^2=8/49t^2#

#color(brown)("c)")# Power, #P= (dW)/dt#
#P(t)= 8/49 (d(t^2))/dt= 16/49t# Evaluate power, #P(t)|_(t=1)=P(1)#
#P(1) = 16/49J/s= 16/49 "Watts"#

Apr 12, 2016

Answer:

#=3.92Js^-1#, rounded to second decimal place.

Explanation:

For constant acceleration, the kinematic expression between velocity, acceleration and time is
#v=u+at#, Inserting given values and solving for acceleration
#4=0+axx7#
or #a=4/7ms^-2#
We know that Force #F=ma#
or #F=12xx4/7=6.86N# rounded to two decimal places.
It is given that during the period acceleration is constant.
Implies that force is also constant during the period.

Let the object move through a distance #dx# in time #dt#
Work done #dW=vecF cdotvecds#, if force and distance moved are in the same direction
Power applied #(dw)/(dt)=Fcdot(ds)/dt#
#=Fcdot at#, #( :. (ds)/dt=v)#
Power applied at #t=1#
#=Fcdot at|_(t=1)#

#P|_(t=1)=6.86xx4/7xx1#
#=3.92Js^-1# rounded to second decimal place.