EJU - Examination for Japanese University Physics: Mechanics and Energy Questions and Answers — Questions and Answers
Question 1: A 2 kg object, initially at rest on a frictionless horizontal surface, is subjected to a constant horizontal force of 10 N. What is the kinetic energy of the object after it has been displaced by 4 m?
- 20 J
- 80 J
- 40 J (Correct answer)
- 10 J
Correct answer: 40 J
According to the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy (W_net = ΔKE). The work done (W) by the constant force (F) over a displacement (d) is calculated as W = F × d. Here, W = 10 N × 4 m = 40 J. Since the object starts from rest, its initial kinetic energy is zero. Therefore, the final kinetic energy is equal to the work done, which is 40 J.
Question 2: A simple pendulum with a 0.5 kg bob is released from rest at a height 'h' above its lowest point. If its speed at the lowest point is 2 m/s, what is the initial height 'h'? (Assume g = 10 m/s² and neglect air resistance).
- 0.4 m
- 0.1 m
- 0.5 m
- 0.2 m (Correct answer)
Correct answer: 0.2 m
This problem uses the principle of conservation of mechanical energy. The potential energy (PE) at the highest point is completely converted into kinetic energy (KE) at the lowest point. So, PE_initial = KE_final. The formula for potential energy is PE = mgh, and for kinetic energy is KE = ½mv². Setting them equal: mgh = ½mv². The mass 'm' cancels out, leaving gh = ½v². Solving for h: h = v² / (2g). Plugging in the values: h = (2 m/s)² / (2 × 10 m/s²) = 4 / 20 = 0.2 m.
Question 3: A 4 kg cart moving at 5 m/s collides with a 6 kg cart that is initially at rest. The two carts stick together after the collision. What is their common velocity after the collision?
- 3 m/s
- 5 m/s
- 2 m/s (Correct answer)
- 2.5 m/s
Correct answer: 2 m/s
In a perfectly inelastic collision where objects stick together, total momentum is conserved. The total momentum before the collision equals the total momentum after. The formula is m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)v_f. Before the collision: p_initial = (4 kg)(5 m/s) + (6 kg)(0 m/s) = 20 kg·m/s. After the collision: p_final = (4 kg + 6 kg)v_f = (10 kg)v_f. Setting p_initial = p_final: 20 kg·m/s = (10 kg)v_f. Solving for v_f gives v_f = 20 / 10 = 2 m/s.
Question 4: An object is launched with an initial velocity at an angle to the horizontal. Neglecting air resistance, which of the following quantities is zero when the object reaches the highest point of its trajectory?
- The horizontal component of its velocity
- The vertical component of its velocity (Correct answer)
- Its acceleration
- The horizontal component of its acceleration
Correct answer: The vertical component of its velocity
In projectile motion, the velocity has two components: horizontal and vertical. The horizontal velocity is constant (assuming no air resistance). The vertical velocity is affected by gravity, decreasing as the object rises and increasing as it falls. At the very peak of its trajectory, the object momentarily stops moving upward before it begins to fall, so its vertical velocity is zero at that instant. Its acceleration is never zero; it is always equal to the acceleration due to gravity (g), which acts downwards.
Question 5: A 5 kg box is pulled across a horizontal floor by a force of 40 N. The coefficient of kinetic friction between the box and the floor is 0.3. What is the net work done on the box if it is moved a distance of 10 m? (Use g = 10 m/s²).
- 400 J
- 150 J
- 550 J
- 250 J (Correct answer)
Correct answer: 250 J
The net work done is the work done by the net force. First, calculate the forces. The applied force is F_app = 40 N. The force of friction is f_k = μ_k * N, where N is the normal force. On a horizontal surface, N = mg = 5 kg * 10 m/s² = 50 N. So, f_k = 0.3 * 50 N = 15 N. The net force is F_net = F_app - f_k = 40 N - 15 N = 25 N. The net work is W_net = F_net * distance = 25 N * 10 m = 250 J.
Question 6: For an object undergoing uniform circular motion at a constant speed, which of the following statements is correct?
- Its velocity is constant.
- Its acceleration is directed tangent to the circle.
- Its acceleration is zero because its speed is constant.
- Its acceleration vector is directed towards the center of the circle. (Correct answer)
Correct answer: Its acceleration vector is directed towards the center of the circle.
In uniform circular motion, the speed is constant, but the velocity is not, because velocity is a vector quantity that includes direction, and the direction is continuously changing. This change in velocity means there is an acceleration. This acceleration, called centripetal acceleration, is always directed radially inward, towards the center of the circle, perpendicular to the object's velocity vector (which is tangent to the circle).
EJU - Examination for Japanese University Physics: Mechanics and Energy Questions and Answers
A 2 kg object, initially at rest on a frictionless horizontal surface, is subjected to a constant horizontal force of 10 N.
What is the kinetic energy of the object after it has been displaced by 4 m?