Amara is sitting on her sled. Her and the sled have a combined mass of $50.5\text{ kg}$. While on top of the hill, she is pushed with a constant force of $F_\text{applied}$ at an angle of $\theta = 41.2^\circ$ above the horizontal for $11.4\text{ s}$ until she reaches a speed of $5.1\text{ m/s}$ at the edge of the hill. If the coefficient of kinetic friction between Amara's sled and the snow is $\mu = 0.05$, what is the magnitude of $F_\text{applied}$?
We can use the principle of work and energy to solve this problem. The work done by the applied force is equal to the change in the sled's kinetic energy:
�applied=Δ�Wapplied=ΔK
where $W_\text{applied}$ is the work done by the applied force, and $\Delta K$ is the change in kinetic energy. The work done by the applied force is given by:
�applied=�applied�cos�Wapplied=Fapplieddcosθ
where $d$ is the distance traveled by Amara and her sled. We can calculate $d$ using the formula for the distance traveled with constant acceleration:
�=12��2d=21at2
where $a$ is the acceleration and $t$ is the time. The acceleration of Amara and her sled is given by:
�=�appliedcos�−�friction�a=mFappliedcosθ−ffriction
where $f_\text{friction}$ is the force of friction and $m$ is the mass of Amara and her sled. The force of friction is given by:
�friction=��ffriction=μN
where $N$ is the normal force, which is equal to the weight of Amara and her sled:
�=��N=mg
where $g$ is the acceleration due to gravity. Putting it all together, we have:
\begin{align*} d &= \frac{1}{2}at^2 \ &= \frac{1}{2} \frac{F_\text{applied} \cos\theta - f_\text{friction}}{m} t^2 \ &= \frac{1}{2} \frac{F_\text{applied} \cos\theta - \mu mg}{m} t^2 \ \end{align*}
The final speed of Amara and her sled is given by:
��2=��2+2�Δ�vf2=vi2+2aΔx
where $v_i$ is the initial velocity (which is zero), $\Delta x$ is the distance traveled, and $v_f$ is the final velocity. Solving for $\Delta x$, we get:
Δ�=��22�Δx=2avf2
Putting it all together, we have:
\begin{align*} W_\text{applied} &= \Delta K \ F_\text{applied} d \cos\theta &= \frac{1}{2} m v_f^2 \ F_\text{applied} \frac{1}{2} \frac{F_\text{applied} \cos\theta - \mu mg}{m} t^2 \cos\theta &= \frac{1}{2} m v_f^2 \ F_\text{applied} \frac{1}{2} \left(F_\text{applied} \cos^2\theta - \mu mg \cos\theta\right) t^2 &= \frac{1}{2} m v_f^2 \ F_\text{applied} &= \frac{m v_f^2}{t^2 \left(\frac{1}{2} \cos^2\theta - \frac{\mu g}{2}\cos\theta\right)} \end{align*}
Substituting the given values, we get:
F_{applied} = \frac{(50.5\text{ m/s})(5.1 \text{ s}) = 9.9 \text{ N}.
