An object with a mass of #15 kg# is moving at #9 m/s# over a surface with a kinetic friction coefficient of #2#. How much power will it take to accelerate the object at #3 m/s^2?
1 Answer
Explanation:
Power is defined as the rate of transfer of energy, which can be stated mathematically as follows:
#"Power"=(dE_("system"))/(dt)# where
#E_"system"# is the energy of the system and#t# is time
The above statement depends on the derivative of the system's energy with respect to time. We can also express this as the average power, similar to the way we have with velocity and average velocity.
#color(purple)("Average Power"=(DeltaE_("system"))/(Deltat))#
A convenient expression for the average power in our case is:
#color(crimson)(P_"avg"=F*v)# where
#F# is the applied force and#v# is the velocity
Therefore, if we calculate the applied force on the object, we can find the power required to cause the given acceleration.
Here is an inventory of the given information:
#|->"m"=15"kg"# #|->"v"_i=9"m"//"s"# #|->mu_k=2# #|->a=3"m"//"s"^2#
Taking inventory of the forces acting on the object, we have:
#sumF_x=F_a-f_k=ma# Where
#F_a# is the applied force and#f_k# is the force of kinetic friction
#sumF_y=n-F_G=0# Where
#n# is the normal force and#F_G# is the force of gravity. Note that the net force is zero because the object does not move vertically (dynamic equilibrium).
Therefore, we have the following equations:
#color(skyblue)(F_a=ma+f_k)#
#color(skyblue)(n=F_G=mg)#
We also know that the equation for kinetic friction is given by:
#f_k=mu_kn#
So, we have:
#F_a=ma+mu_kn#
#=color(crimson)(ma+mu_kmg)#
Therefore, we now have the following equation for power:
#P=(ma+mu_kmg)v#
which we have all of the necessary information to solve.
#P=[(15"kg")(3"m"//"s"^2)+(2)(15"kg")(9.81"m"//"s"^2)] (9"m"//"s"^2)#
#=3053.7"W"#
#~~3054"W"#
