The figure shows three forces applied to a trunk that moves leftward by 3.34 m over a frictionless floor. The force magnitudes are F1 = 5.39 N, F2 = 8.74 N, and F3 = 2.52 N, and the indicated angle is θ = 60°. ?

(a) During the displacement, what is the net work done on the trunk by the three applied forces, the gravitational force, and the normal force? (b) Is there a net transfer of energy to or from the trunk? (c) Does the kinetic energy of the trunk increase or decrease?enter image source here

1 Answer
Nov 30, 2015

Answer:

  • Net work is #3.4068J#, Gravitational force #=#Weight#=5.04906N# , normal force equals to zero .
  • Net transfer is to the trunk.
  • Kinetic energy increases .

Explanation:

Part A
Work
#=vecF_"net"*vecs#
#=(-5.39N+8.74cos60^oN)(-3.34m)#
#=3.4068J#

Taking the left directions as negative and right as positive, we want the sum of the forces that are pointing horizontally. Hence, #8.74cos60^oN# plus #-5.39N#.

Gravitational force (weight)
#=8.74sin60^oN-2.52N-W=0#
Hence #W=5.04906N#

Since the trunk is not moving up or down, the forces pulling up and down must cancel each other.
Forces up#=8.74sin60^oN#
Forces down#=2.52N# plus the weight of the trunk, #W#.

Normal force is zero because the question says frictionless floor. For that to be possible, the normal force must be zero. See coefficient of static friction .

#F_"static friction"=muN#. If #F_"static friction"=0#, #N=0#.

Part B
Energy transferred to the trunk because there is work done on the trunk (positive value).

Part C
Kinetic energy increases because you have positive work .
Other way to look at it is the net force is to the left, when there is a net force, there is acceleration .
Faster you go, higher your kinetic energy.
#F_"net"=ma#

Cheers.