Chapter 6, Problem 33P

A sled is initially given a shove up a frictionless 23.0° incline. It reaches a maximum vertical height 1.22 m higher than where it started at the bottom. What was its initial speed?

To determine

Initial speed of the sled.

Answer to Problem 33P

Solution:Initial speed of the sled is 4.9 m/s.

Explanation of Solution

Given:Final height, $h_f=1.22\text{ m}$

Formula Used:

The forces acting on sled are the force due to gravity and normal force. Work done by the normal force is zero because of perpendicular to the direction motion.

According to the law of conservation of energy

  $\begin{array}{l}{E}_{initial}=E_{final}\\ mg{h}_i+\frac{1}{2}m{v}_1{}^2=mg{h}_f+\frac{1}{2}m{v}_f{}^2\end{array}$

Where m is the mass of sled, $h_i$ is the initial height of sled, $v_i$ is the initial velocity of the sled, $v_f$ is the final velocity of the sled, $h_f$ is the final height of the sled, g is acceleration due to gravity.

Calculations:

The sled moving along the ground before it has gained any height. It has kinetic energy and no potential energy. The sled having slid up the plane and come to a rest, it has potential energy and no kinetic energy,

Substituting final height value in above equation and solve for initial velocity

  $\begin{array}{l}mg{h}_i+\frac{1}{2}m{v}_i{}^2=mg{h}_f+\frac{1}{2}m{v}_f{}^2\\ 0+\frac{1}{2}m{v}_i{}^2=mg{h}_f+0\\ \frac{1}{2}m{v}_i{}^2=mg{h}_f\\ v_i{}^2=2g{h}_f\\ v_i=\sqrt{2g{h}_f}\\ v_i=\sqrt{2\times 9.8\times 1.22}\\ v_i=\sqrt{23.91}\\ v_i=4.88=4.9\text{ m/s}\end{array}$

The initial velocity of the sled is $4.9\text{ m/s}$ and angle of the incline is not relevant to the solution.

Conclusion:The sled is moving atthe initial speed of $4.9\text{ m/s}$ to reach the maximum vertical height of $1.22\text{m}$ higher than where it started at the bottom.