How much work would it take to push a # 1 kg # weight up a # 6 m # plane that is at an incline of # pi / 3 #?

1 Answer
Aug 20, 2017

Answer:

#W~~51"J"#

Explanation:

Work done by a constant force is given as the dot product of the force and displacement vectors:

#color(blue)(W=Fdcostheta)#

  • Where #theta# is the angle between the force and displacement vectors. This is not the angle of incline.

Diagram:

enter image source here

  • Where #vecF_A# is the applied (pushing) force, #vecn# is the normal force, and #vecF_g# is the force of gravity, broken up into its parallel (x, horizontal) and perpendicular (y, vertical) components.

  • I will define up the incline as positive

We can use Newton's second law to determine the net force on the object.

#color(blue)(F_(x" net")=sumF_x=F_A-F_(Gx)=ma_x)#

#color(blue)(F_(y" net")=sumF_x=n-F_(Gy)=ma_y)#

  • We will assume dynamic equilibrium, as it is assumed that we are looking for the minimum amount of work required to push the object up the incline.

#=>F_A-F_(Gx)=0#

#=>n-F_(Gy)=0#

Which gives:

#=>color(blue)(F_A=F_(Gx))#

#=>color(blue)(n=F_(Gy))#

Therefore, the magnitude of the pushing force required is equal to the parallel component of the force of gravity. We can find this component using our diagram and basic trigonometry.

#=>sin(theta)="opposite"/"hypotenuse"#

#=>sin(theta)=F_(Gx)/F_G#

#=>F_(Gx)=F_Gsin(theta)#

And as we know that #F_G=mg#:

#=>color(blue)(F_(Gx)=mgsin(theta))#

Finally, we have:

#F_A=mgsin(theta)#

Putting this into the equation for work:

#W=mgsin(theta_i)dcos(theta)#

Since the pushing force is in the same direction as the displacement (up the ramp), #theta=0#, and so we have:

#W=mgsin(theta)d#

We have the following information:

  • #|->m=1"kg"#
  • #|->d=6"m"#
  • #|->theta_i=pi/3#
  • #|->g=9.81"m"//"s"^2#

#=>W=(1"kg")(9.81"m"//"s"^2)(6"m")sin(pi/3)#

#=>color(blue)(W~~51"J")#