Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 4: Q4-91GP (page 75)
A 72-kg water skier is being accelerated by a ski boat on a flat (鈥済lassy鈥�) lake. The coefficient of kinetic friction between the skier鈥檚 skis and the water surface is \({\mu _k} = 0.25\) (Fig. 4鈥�74). (a) What is the skier鈥檚 acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude \({F_T} = 240\;{\rm{N}}\) to the skier \(\left( {\theta = 0^\circ } \right)\)? (b) What is the skier鈥檚 horizontal acceleration if the rope pulling the skier exerts a force of \({F_T} = 240\;{\rm{N}}\) on the skier at an upward angle \(\theta = 12^\circ \)? (c) Explain why the skier鈥檚 acceleration in part (b) is greater than that in part (a).
Short Answer
The obtained values of acceleration in parts (a) and (b) are \(a = 0.88\;{\rm{m/}}{{\rm{s}}^2}\) and \(a = 0.9\;{\rm{m/}}{{\rm{s}}^2}\), respectively.
Step by step solution
Calculate the forces in vertical and horizontal directions
Draw the free body diagram for the skier in part (a) at zero upward angle and in part (b) at 12-degree upward angle. In determining the acceleration, use Newton鈥檚 second law for horizontal and vertical directions.
Given data:
The mass of the water skier is\(m = 72\;{\rm{kg}}\).
The kinetic coefficient of friction between the skier and the water is\({\mu _k} = 0.25\).
The magnitude of tension force is\({F_T} = 240\;{\rm{N}}\).
The upward angle made by the skier is\(\theta = 12^\circ \).
The free body diagram of the skier is as follows:
Calculate the acceleration of the skier
The relation to calculate the forces in the horizontal direction is given by:
\(\begin{aligned}\Sigma {F_x} &= ma\\{F_T} - {F_{fr}} &= ma\\{F_T} - {\mu _k}N &= ma\end{aligned}\)
Here, a is the acceleration of the skier.
On plugging the values in the above relation, you get:
\(\begin{aligned}{F_T} - {\mu _k}mg &= ma\\\left( {240\;{\rm{N}}} \right) - \left( {0.25 \times 72\;{\rm{kg}} \times {\rm{9}}{\rm{.81}}\;{\rm{m/}}{{\rm{s}}^2}} \right) &= \left( {72\;{\rm{kg}}} \right)a\\a &= 0.88\;{\rm{m/}}{{\rm{s}}^2}\end{aligned}\)
Thus, \(a = 0.88\;{\rm{m/}}{{\rm{s}}^2}\) is the acceleration of the skier.
Calculate the forces in vertical and horizontal directions
The free body diagram of the skier is as follows:
The relation to calculate the forces in the vertical direction is given by:
\(\begin{aligned}\Sigma {F_y} &= 0\\N - W &= 0\\N &= mg\end{aligned}\)
Here, \(N\) is the normal force, \(W\) is the weight, and g is the gravitational acceleration.
Calculate the acceleration of the system
On plugging the values in the above relation, you get:
\(\begin{aligned}{F_T}\cos \theta - {\mu _k}\left( {mg - {F_T}\sin \theta } \right) &= ma\\a &= \left( {\frac{{{F_T}\left( {\cos \theta + {\mu _k}\sin \theta } \right) - {\mu _k}mg}}{m}} \right)\\a &= \left( {\frac{{\left( {240\;{\rm{N}}} \right)\left( {\cos 12^\circ + \left( {0.25} \right)\sin 12^\circ } \right) - \left( {0.25} \right)\left( {72\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{\left( {72\;{\rm{kg}}} \right)}}} \right)\\a &= 0.9\;{\rm{m/}}{{\rm{s}}^2}\end{aligned}\)
Thus, \(a = 0.9\;{\rm{m/}}{{\rm{s}}^2}\) is the acceleration of the system.
Calculate the acceleration of the system
In pat (b), the value of the acceleration is higher than part (a) because the upward tilt of the tow rope decreases the amount of frictional force by reducing the normal force.
Thus, the acceleration in part (b) is greater than acceleration in part (a).
Over 30 million students worldwide already upgrade their learning with 91影视!
