Unit: 3. Work , Energy and Power
Weightage : 10-15%
A child of unknown mass is on a swing of unknown length that varies in height from \(75\,\mathrm{cm}\) at its lowest height above the ground to a maximum height of \(225\,\mathrm{cm}\) above the ground. Is there enough information to find the speed of the swing at its lowest point?
Correct Answer: \( \boxed{\mathrm{C}} \)
The speed at the lowest point can be found using conservation of mechanical energy. The mass of the child is not needed because it cancels out.
The change in height is
\( \Delta h = 2.25\,\mathrm{m}-0.75\,\mathrm{m}=1.50\,\mathrm{m} \)
Applying conservation of energy:
\( m g \Delta h=\frac{1}{2}mv^2 \)
Canceling \(m\),
\( g\Delta h=\frac{1}{2}v^2 \)
\( v=\sqrt{2g\Delta h} \)
\( v=\sqrt{2(10)(1.50)} \)
\( v=\sqrt{30} \)
\( v\approx 5.5\,\mathrm{m/s} \)
Therefore, there is sufficient information to determine the speed, and the speed at the lowest point is approximately \(5.5\,\mathrm{m/s}\).
Hence, the correct answer is \( \boxed{\mathrm{C}} \).
A rubber ball with mass \( \mathrm{0.20\ kg} \) is dropped vertically from a height of \( \mathrm{1.5\ m} \) above a floor.
The ball bounces off the floor, and during the bounce \( \mathrm{0.60\ J} \) of energy is dissipated.
What is the maximum height of the ball after the bounce?
Correct Answer: \( \boxed{\mathrm{C}} \)
Before striking the floor, the ball’s gravitational potential energy is
\( \mathrm{U_i=mgh} \)
\( \mathrm{U_i=(0.20)(10)(1.5)=3.0\ J} \)
When the ball bounces, \( \mathrm{0.60\ J} \) of energy is dissipated.
Therefore, the energy remaining after the bounce is
\( \mathrm{E=3.0-0.60=2.40\ J} \)
At the maximum height after the bounce, all of this energy is gravitational potential energy:
\( \mathrm{mgh_f=2.40} \)
Substituting \( \mathrm{m=0.20\ kg} \) and \( \mathrm{g=10\ m/s^2} \):
\( \mathrm{(0.20)(10)h_f=2.40} \)
\( \mathrm{2h_f=2.40} \)
\( \mathrm{h_f=1.2\ m} \)
Therefore, the maximum height reached after the bounce is
\( \boxed{\mathrm{1.2\ m}} \)
Hence, the correct answer is \( \boxed{\mathrm{C}} \).
A ball is dropped from rest and falls to the floor. The initial gravitational potential energy of the ball-Earth-floor system is \( \mathrm{10\ J} \).
The ball then bounces back up to a height where the gravitational potential energy is \( \mathrm{7\ J} \).
What was the mechanical energy of the ball-Earth-floor system the instant the ball left the floor?
Correct Answer: \( \boxed{\mathrm{C}} \)
Initially, the ball-Earth-floor system has gravitational potential energy
\( \mathrm{U_i=10\ J} \)
As the ball falls, this potential energy is converted into kinetic energy.
During the collision with the floor, some mechanical energy is lost due to deformation, sound, and thermal energy.
After the bounce, the ball rises to a height where its gravitational potential energy is
\( \mathrm{U_f=7\ J} \)
At the highest point of the rebound, the ball’s speed is zero, so
\( \mathrm{K=0} \)
Therefore, the total mechanical energy at the top of the bounce is
\( \mathrm{E_{mech}=7\ J} \)
Once the ball leaves the floor, only gravity acts on it, so mechanical energy is conserved during the upward motion.
Thus, the mechanical energy immediately after leaving the floor must be the same as the mechanical energy at the top of the bounce:
\( \mathrm{E_{mech}=7\ J} \)
Hence, the mechanical energy of the ball-Earth-floor system the instant the ball left the floor was
\( \boxed{\mathrm{7\ J}} \)
Therefore, the correct answer is \( \boxed{\mathrm{C}} \).
