Question

Susan’s $10$ kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled $30°$above the floor. The tension is a constant $30$N and the coefficient of friction is $0.20$. Use work and energy to find Paul’s speed after being pulled$3.0$ m.

Short answer

Paul's speed after being pulled $3$m is $2.37$m/s

Step-by-step solution

Step 1. Given Information

Mass $m=10$kg

Tension $=30$N

the coefficient of friction $=0.20$

Step 2. Find the magnitude of friction

We use second Newton's law for direction perpendicular to motion, or$y$ direction

$$\begin{gathered} m{a}_y=0=-F_{gravitational}+N+F\sin 30° \\ N=mg-F\sin 30° \\ N=10kg·9.8m/s^2-30N·0.5 \\ N=83N \end{gathered}$$

Friction is defined as

$$\begin{gathered} F_{friction}=N\mu \\ {F}_{friction}=83N·0.2 \\ {F}_{friction}=16.6N \end{gathered}$$

Step 3. Work done by friction if Paul is pulled 3 m

$$\begin{gathered} W_{friction}=F_{friction}·d \\ {W}_{friction}=16.6N·3m \\ {W}_{friction}=49.8J \end{gathered}$$

Step 4. Work done by rope on Paul

$$\begin{gathered} W_{rope}=F\cos 30°·d \\ {W}_{rope}=30N·\cos 30°·3m \\ {W}_{rope}=77.9J \end{gathered}$$

Step 5. Total work done on mat and Paul

$$\begin{gathered} W_{rope}=W_{rope}+W_{friction} \\ W=77.9J-49.8J \\ W=28.1J \end{gathered}$$

Step 6. Find Paul's speed

From the work-kinetic theorem, we can calculate Paul's speed

$$\begin{gathered} W=∆K \\ W=K_{final}-K_{initial} \\ W=\frac{1}{2}m{v^2}_{final} \\ \Rightarrow v_{final}=\sqrt{\frac{2W}{m}} \\ {v}_{final}=\sqrt{\frac{2·28.1J}{10kg}} \\ {v}_{final}=2.37m/s \end{gathered}$$