Question
A particle of mass 0.4 kg is released from rest at a height of 1.8 m above the surface of the water in a
tank. There is no instantaneous change of speed when the particle enters the water. The water exerts an upward force of 5.6 N on the particle when it is in the water.
(i) Find the velocity of the particle at the instant when it reaches the surface of the water.
(ii) Find the time that it takes from the instant when the particle enters the water until it comes to instantaneous rest in the water. You may assume that the tank is deep enough so that the particle does not reach the bottom of the tank.
(iii) Sketch a velocity-time graph for the motion of the particle from the instant at which it is released
until it comes to instantaneous rest in the water.
▶️Answer/Explanation
(i) Velocity just before entering the water:
The potential energy lost by the particle is converted into kinetic energy just before entering the water. The potential energy lost is given by \( mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.
\( \text{Potential energy lost} = mgh \)
This potential energy is converted into kinetic energy just before entering the water:
\( \frac{1}{2}m\nu^2 = mgh \)
Solving for \( \nu \):
\( \nu = \sqrt{2gh} \)
Substitute \( g = 9.8 \, \text{m/s}^2 \) and \( h = 1.8 \, \text{m} \):
\( \nu = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 1.8 \, \text{m}} \)
\( \nu = 6 \, \text{m/s} \)
(ii) Time taken to come to rest in the water:
Once in the water, the net force acting on the particle is the difference between the buoyant force (\(5.6 \, \text{N}\)) and its weight (\(mg\)). The net force is then used to find the deceleration.
\( F_{\text{net}} = 5.6 \, \text{N} – 0.4 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)
Calculate \( F_{\text{net}} \):
\( F_{\text{net}} = -3.6 \, \text{N} \)
\( a = \frac{F_{\text{net}}}{m} = -4 \, \text{m/s}^2 \)
Now, we’ll use the kinematic equation \( \nu = u + at \), where \( u \) is the initial velocity (velocity just before entering water), \( a \) is the deceleration, and \( t \) is the time.
\( \nu = 0 \) (since the particle comes to rest)
Solve for \( t \):
\( 0 = \sqrt{2gh} + at \)
\( t = 1.5 \, \text{s} \)
(iii) Velocity-time graph:
The graph starts at the origin (0,0) with a positive gradient, indicating a linear increase in velocity during free fall. After entering the water, there’s a second straight line with a negative gradient, indicating a linear decrease in velocity until the particle comes to rest. The graph finishes at (2.1, 0), showing that the velocity becomes zero at \( t = 2.1 \) seconds.
Question
An elevator moves vertically, supported by a cable. The diagram shows a velocity-time graph which models the motion of the elevator. The graph consists of 7 straight line segments.
The elevator accelerates upwards from rest to a speed of \(2ms^{-1}\) over a period of 1.5s and then travels at this speed for before decelerating to rest over a period of 1 s.
The elevator then remains at rest for 6 s. before accelerating to a speed of \( Vms^{-1}\) downwards over a period of 2s. The elevator travels at this speed for a period of 5s, before decelerating to rest over a period of 1.5s.
(a) Find the acceleration of the elevator during the first 1.5s.
(b) Given that the elevator starts and finishes its journey on the ground floor, find V.
(c) The combined weight of the elevator and passengers on its upward journey is 1500kg. Assuming that there is no resistance to motion, find the tension in the elevator cable on its upward journey when the elevator is decelerating.
▶️Answer/Explanation
(a) Acceleration during the first \(1.5 \, \text{s}\):
The formula for acceleration is \(a = \frac{\Delta v}{\Delta t}\), where \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time.
In the given velocity-time graph, during the first \(1.5 \, \text{s}\), the elevator accelerates from rest to a speed of \(2 \, \text{m/s}\). Therefore, the change in velocity \(\Delta v\) is \(2 \, \text{m/s}\), and the change in time \(\Delta t\) is \(1.5 \, \text{s}\). Plug these values into the formula to find the acceleration \(a\):
\( a = \frac{2 \, \text{m/s}}{1.5 \, \text{s}} = \frac{4}{3} \, \text{m/s}^2 \)
(b) Find \(V\):
For this part, we consider the downward journey. The equation of motion for constant acceleration is given by:
\( s = ut + \frac{1}{2} a t^2 \)
For the first segment (acceleration), \(s_1 = \frac{1}{2} a t_1^2\) where \(a = \frac{4}{3} \, \text{m/s}^2\) and \(t_1 = 1.5 \, \text{s}\).
For the second segment (constant velocity), \(s_2 = V t_2\) where \(V\) is the constant velocity and \(t_2 = 1 \, \text{s}\).
For the third segment (deceleration), \(s_3 = \frac{1}{2} a t_3^2\) where \(a = -2 \, \text{m/s}^2\) and \(t_3 = 1.5 \, \text{s}\)
The total displacement is given by \(s_1 + s_2 + s_3\) and is set equal to zero since the elevator starts and finishes at the same position. Substituting the known values into the equation and solving for \(V\) gives:
\( \frac{1}{2} \left(\frac{4}{3}\right) \times (1.5)^2 + V \times 1 + \frac{1}{2} \times (-2) \times (1.5)^2 = 0 \)
Solving this equation yields \(V = 1.70 \, \text{m/s}\) (rounded to three significant figures).
(c) Tension during upward journey when decelerating:
Use Newton’s second law \(\Sigma F = ma\). The net force is \(T – mg\), where \(T\) is the tension, \(m\) is the mass of the elevator and passengers (\(1500 \, \text{kg}\)), \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)), and \(a\) is the acceleration during deceleration (\(-2 \, \text{m/s}^2\)).
\( T – mg = ma \)
\( T – 1500 \times 9.8 = 1500 \times (-2) \)
Solving this equation gives \(T = 12000 \, \text{N}\).