Question

A 35kg box is pushed 14m by a force of 800n. The coefficient of kinetic friction is 0.4. How much work into accelerating?

Answer 1

Given data:

* The force applied on the box is,

`F_a=800\text{ N}`

* The mass of the box is m = 35 kg.

* The distance traveled by the box is d = 14 m.

* The coefficient of kinetic friction is,

`\mu_k=0.4`

Solution:

The normal force acting on the box is,

`F_N=mg`

where g is the acceleration due to gravity,

Substituting the known values,

`\begin{gathered} F_N=35*9.8 \\ F_N=343\text{ N} \end{gathered}`

The kinetic frictional force acting on the box is,

`F_k=\mu_kF_N`

Substituting the known values,

`\begin{gathered} F_k=0.4*343 \\ F_k=137.2\text{ N} \end{gathered}`

The net force acting on the box in terms of applied force and frictional force is,

`\begin{gathered} F_{\text{net}}=F_a-F_k \\ F_{\text{net}}=800-137.2 \\ F_{\text{net}}=662.8\text{ N} \end{gathered}`

Thus, the work done in accelerating the box is,

`\begin{gathered} W=F_{\text{net}}* d \\ W=662.8*14 \\ W=9279.2\text{ J} \end{gathered}`

Thus, the work done in accelerating the box is 9279.2 J or approximately 9.3 kJ.