91Ó°ÊÓ

When non-conservative forces act on an object, then the work done by these non-conservative forces can be calculated using the work-energy principle.

According to the work-energy principle, work done by the non-conservative forces acting on an object is equal to the total change in the kinetic and potential energies of the object,i.e.,\(W = \Delta KE + \Delta PE\).

Mass of the sled is\(m = 18\;{\rm{kg}}\).

The initial speed of the sled at the bottom of the incline is \(u = 2.3\;{\rm{m/s}}\).

The coefficient of kinetic friction is \({\mu _{\rm{k}}} = 0.25\).

The inclination of the inclined path is \(\theta = 28^\circ \).

Let the sled travels a distance x up on the incline. The free-body diagram of the sled at any instant is shown in the figure below:

The final speed of the sled after traveling the distance xwill be zero, i.e., \(v = 0\;{\rm{m }}{{\rm{s}}^{ - 1}}\).

So, the change in kinetic energy of the sled in traveling a distance x on the incline is as follows:

\(\begin{aligned}\Delta KE &= \frac{1}{2}m\left( {{v^2} - {u^2}} \right)\\ &= - \frac{1}{2}m{u^2}\end{aligned}\)

Height (h) of the sled when it has traveled a distancex on the incline is calculated as follows:

\[\begin{aligned}\sin \theta &= \frac{h}{x}\\h &= x\sin \theta \end{aligned}\]

The initial potential energy of the sled at the bottom of the incline is zero. Thus, the change in potential energy of the sled in traveling a distance x on the incline is as follows:

\(\begin{aligned}\Delta PE &= P{E_f} - P{E_i}\\ &= mgh - 0\\ &= mgx\sin \theta \end{aligned}\)\(\)

The frictional force acting on the sled is given as:

\(\begin{aligned}F &= {\mu _{\rm{k}}}N\\ &= {\mu _{\rm{k}}}mg\cos \theta \end{aligned}\)

Here, N is the normal reaction on the sled,equal to the vertical component of weight acting downwards, as shown in the figure above.

According to the work-energy principle, work done by the frictional force acting on the sled is given as follows:

\(\begin{aligned}W &= \Delta KE + \Delta PE\\ - Fx &= - \frac{1}{2}m{u^2} + \left( {mg{\rm{xsin}}\theta } \right)\\\left( {{\mu _{\rm{k}}}mg\cos \theta } \right)x &= \frac{1}{2}m{u^2} - \left( {mg{\rm{xsin}}\theta } \right)\\x &= \frac{{{u^2}}}{{2g\left( {{\mu _{\rm{k}}}\cos \theta + \sin \theta } \right)}}\end{aligned}\)

On substituting given values in the above equation,

\(\begin{aligned}x &= \frac{{{u^2}}}{{2g\left( {\mu \cos \theta + \sin \theta } \right)}}\\ &= \frac{{{{\left( {2.3\;{\rm{m/s}}} \right)}^2}}}{{2 \times \left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {0.25\cos 28^\circ + \sin 28^\circ } \right)}}\\ &= 0.39\;{\rm{m}}\end{aligned}\)

Thus, the sled will travel 0.39 m up on the incline.

The sled will not be stuck on an incline after traveling a distance x if it travels back downwards along the incline.

Thus, while returning, the force of friction will act in the opposite direction, i.e., along upwards parallel to the inclination. For this downward motion to become possible, frictional force must be less than the horizontal component of weight acting downwards parallel to the inclination, i.e.,

\(\begin{aligned}F &< mg\sin \theta \\{\mu _{\rm{k}}}mg\cos \theta &< mg\sin \theta \\{\mu _{\rm{k}}} &< \tan \theta \\ &< \tan 28^\circ \\ &< 0.53\end{aligned}\)

Thus, the coefficient of kinetic friction must be less than 0.53 so that sled does not get stuck after traveling a distance of 0.39 m up along the incline.

The sled slides back downwards along the incline. The final velocity of the sled when it returns to its starting point is \(v'\).

Since the sled moves from the bottom of the incline and travels up a distance of x m, then returns to its initial point, the total distance traveled by the sled is 2x. Also, the change in the potential energy of the sled is zero.

The change in kinetic energy of the sled is as follows:

\(\Delta KE = \frac{1}{2}m{v'^2} - \frac{1}{2}m{u^2}\)

Applying the work-energy principle on the motion of sled,

\(\begin{aligned}W &= \Delta KE + \Delta PE\\ - F\left( {2x} \right) &= \frac{1}{2}m{{v'}^2} - \frac{1}{2}m{u^2} + 0\\ - \left( {{\mu _{\rm{k}}}mg\cos \theta } \right)2x &= \frac{1}{2}m{{v'}^2} - \frac{1}{2}m{u^2}\\\frac{1}{2}m{{v'}^2} &= \frac{1}{2}m{u^2} - \left( {{\mu _{\rm{k}}}mg\cos \theta } \right)2x\\{{v'}^2} &= {u^2} - 4x{\mu _{\rm{k}}}mg\cos \theta \end{aligned}\)

On substituting given values in the above equation,

\(\begin{aligned}{{v'}^2} &= {\left( {2.3\;{\rm{m/s}}} \right)^2} - 4\left( {0.39\;{\rm{m}}} \right)\left( {0.25} \right)\left( {18\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\cos 28^\circ \\ &= 5.29 - 3.36\\ &= 1.93\\v' &= 1.4\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of the sled when it returns to its starting point is 1.4 m/s.