Q25E - University Physics with Modern Physics

Question

A sled with mass 12 kg moves in a straight line on a frictionless, horizontal surface. At one point in its path, its speed is 4 m/s; after it has traveled 2.50 m beyond this point, its speed is 6.00 m/s. Use the work–energy theorem to find the force acting on the sled, assuming that this force is constant and that it acts in the direction of the sled’s motion.

Step-by-Step Solution

Answer

The force acting on the sled is 48 N.

Step 1: Identification of given data

The given data can be listed below,

• The mass of the sled is, m =12 kg
• The speed of the sled is, $v_1=4 \mathrm{m}/\mathrm{s}$
• The final speed of the sled is, $v_2=6 \mathrm{m}/\mathrm{s}$
• The distance travelled by the sled is, x =2.50 m

Step 2: Concept/Significance of work energy theorem

A connection between work and energy is shown by the work-energy theorem. It is actually a conservation of energy equation, meaning that energy can only change phases from one kind to another and cannot be generated or destroyed.

Step 3: Determination of the force acting on the sled

The total work done on the sled is given by,

 

$$\begin{gathered} W_{tot}=K_2-K_1 \\ =\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2 \end{gathered}$$                                                                                                    …(i)

 

The work done by the force is given by,

 

$$\begin{gathered} W_{tot}=\overrightarrow{f} .\overrightarrow{s} \\ =fs\cos \varphi \end{gathered}$$                                                                                                                  …(ii)

 

Compare equation (i) and (ii) the force on the sled is given by,

$$\begin{gathered} fs\cos \varphi =\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2 \\ f=\frac{1}{2s}m\left(v_2^2-v_1^2\right) \end{gathered}$$ 

 

Substitute all the values in the above,

$$\begin{gathered} \mathrm{f}=\frac{1}{2\mathrm{s}}\mathrm{m}\left({\mathrm{v}}_2^2-{\mathrm{v}}_1^2\right) \\ =\frac{1}{2\left(2.5 \mathrm{m}\right)}\left(12 \mathrm{m}\right)\left(6^2-4^2\right){\left(\mathrm{m}/\mathrm{s}\right)}^2 \\ =\frac{1}{5}\left(12 \mathrm{kg}\right)\left(36-16\right)\mathrm{m}/{\mathrm{s}}^2 \\ =48 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\left(\frac{1 \mathrm{N}}{1 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =48 \mathrm{N} \end{gathered}$$ 

Thus, the force acting on the sled is 48 N