Question

A sled of mass 26 kg has an 18 kg child on it. If big brother is pulling with a 30 N force to the right and 10 N up, and big sister is pushing with a 40 N force to the right and 16 N down. Assuming that μk = 0.12, calculate:A. The force of frictionB. The acceleration of the sled.

Answer 1

Given data:

* The mass of the sled is m_1 = 26 kg.

* The mass of the child is m_2 = 18 kg.

* The force acting on the sled in the upward direction by the big brother is F_1 = 10 N.

* The force towards the right by the big brother is F_2 = 30 N.

* The force towards the right by the big sister is F_3 = 40 N.

* The force acting on the sled in the downward direction by the big sister is F_4 = 16 N.

* The coefficient of friction is,

`\mu_k=0.12`

Solution:

The net mass on the sled with the child is,

`\begin{gathered} m=m_1+m_2 \\ m=26+18 \\ m=44\text{ kg} \end{gathered}`

The weight of the sled with the child is,

`\begin{gathered} w=mg \\ w=44*9.8 \\ w=431.2\text{ N} \end{gathered}`

The normal force acting on the sled is,

`\begin{gathered} F_N=w+F_4-F_1 \\ F_N=431.2+16-10 \\ F_N=437.2\text{ N} \end{gathered}`

(A). The frictional force acting on the sled is,

`\begin{gathered} F_k=\mu_kF_N \\ F_k=0.12*437.2 \\ F_k=52.464\text{ N} \\ F_k=52.5\text{ N} \end{gathered}`

Thus, the frictional force acting on the sled is 52.5 N.

(B). The force applied on the sled towards the right by the big brother and big sister is,

`\begin{gathered} F=F_2+F_3 \\ F=30+40 \\ F=70\text{ N} \end{gathered}`

The frictional force acts opposite to the direction of applied force, thus, the frictional force acts towards the left.

The net force acting on the sled is,

`\begin{gathered} F_{\text{net}}=F-F_k \\ F_{\text{net}}=70-52.5 \\ F_{\text{net}}=17.5\text{ N} \end{gathered}`

According to newton's second law, the acceleration of the sled is,

`\begin{gathered} F_{\text{net}}=ma \\ a=\frac{F_{\text{net}}}{m} \\ a=(17.5)/(44) \\ a=0.4ms^(-2) \end{gathered}`

Thus, the acceleration of the sled is 0.4 meters per second squared.