Question

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?

Answer 1

• 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.