# 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?

Nov 30, 2015
• Net work is $3.4068 J$, Gravitational force $=$Weight$= 5.04906 N$ , normal force equals to zero .
• Net transfer is to the trunk.
• Kinetic energy increases .

#### Explanation:

Part A
Work
$= {\vec{F}}_{\text{net}} \cdot \vec{s}$
$= \left(- 5.39 N + 8.74 \cos {60}^{o} N\right) \left(- 3.34 m\right)$
$= 3.4068 J$

Taking the left directions as negative and right as positive, we want the sum of the forces that are pointing horizontally. Hence, $8.74 \cos {60}^{o} N$ plus $- 5.39 N$.

Gravitational force (weight)
$= 8.74 \sin {60}^{o} N - 2.52 N - W = 0$
Hence $W = 5.04906 N$

Since the trunk is not moving up or down, the forces pulling up and down must cancel each other.
Forces up$= 8.74 \sin {60}^{o} N$
Forces down$= 2.52 N$ 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 .

${F}_{\text{static friction}} = \mu N$. If ${F}_{\text{static friction}} = 0$, $N = 0$.

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.
${F}_{\text{net}} = m a$

Cheers.