Final answer:
a) The box will start moving with a velocity of 0.1 m/s. b) The work done by friction on the box is 0.098 J. c) The velocity of the box when it reaches the edge of the table is 0.442 m/s. d) The magnitude of the velocity of the box right before it hits the ground is 4.43 m/s.
Step-by-step explanation:
a) To find the velocity at which the box starts moving, we can use the principle of conservation of momentum. The momentum of the bullet before it hits the box is given by m * v, where m is the mass of the bullet and v is its velocity. The momentum of the box and the bullet after the collision can be written as (M + m) * V, where V is the velocity of the box and the bullet together. Since the bullet gets stuck in the box and their velocities become the same, we have:
m * v = (M + m) * V
Solving for V, we get V = (m * v) / (M + m). Substituting the values, we get V = (0.01 kg * 10 m/s) / (1 kg + 0.01 kg) = 0.1 m/s.
Therefore, the box will start moving with a velocity of 0.1 m/s.
b) The work done by friction can be calculated using the formula for work: W = F * d * cos(theta), where F is the force of friction, d is the distance the box moves, and theta is the angle between the direction of the force and the direction of motion. In this case, the force of friction is given by F = mu * N, where mu is the coefficient of friction and N is the normal force. Since the box is moving horizontally, the angle between the force and the direction of motion is 0 degrees, so cos(theta) = 1.
The normal force can be calculated as the weight of the box, which is N = M * g, where g is the acceleration due to gravity. The work done by friction can then be written as W = (mu * M * g) * d.
Substituting the values, we get W = (0.01 * 1 kg * 9.8 m/s^2) * 1 m = 0.098 J.
Therefore, the work done by friction on the box while it moves and reaches the edge of the table is 0.098 J.
c) The velocity of the box when it reaches the edge of the table can be calculated using the equation for linear motion: v^2 = u^2 + 2 * a * s, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. In this case, the box starts from rest, so u = 0. The acceleration can be calculated using Newton's second law: F = m * a, where F is the net force acting on the box. The net force in this case is the force of friction, so F = mu * N. Substituting the values, we get F = 0.01 * 1 kg * 9.8 m/s^2 = 0.098 N.
The displacement of the box can be calculated as the distance from the table to the edge, so s = 1 m.
Substituting the values into the equation, we get v^2 = 0^2 + 2 * (0.098 N / 1 kg) * 1 m = 0.196 m^2/s^2. Taking the square root of both sides, we get v = 0.442 m/s.
Therefore, the velocity of the box when it reaches the edge of the table is 0.442 m/s.
d) The magnitude of the velocity of the box right before it hits the ground can be calculated using the equation for free fall: v^2 = u^2 + 2 * a * s, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and s is the vertical distance the box falls. In this case, the box starts from rest, so u = 0. The acceleration due to gravity is a = 9.8 m/s^2. The vertical distance the box falls can be calculated as the height of the table, so s = 1 m.
Substituting the values into the equation, we get v^2 = 0^2 + 2 * 9.8 m/s^2 * 1 m = 19.6 m^2/s^2. Taking the square root of both sides, we get v = 4.43 m/s.
Therefore, the magnitude of the velocity of the box right before it hits the ground is 4.43 m/s.