A person pushes with a horizontal force of 25 N 4kg block, initially at rest on a horizontal table, a distance of 3m. The coefficient of kinetic friction between the block and table is 0.35.
determine
external work on the block-table system.
the energy dissipated by friction
the kinetic energy of the box fine
the modulus of the velocity of the block
To determine the answers to the given questions, let's break it down step-by-step:
Step 1: Calculate the external work done on the block-table system.
Given that the person pushes the block a distance of 3m with a horizontal force of 25N, the external work done is given by the formula:
Work (W) = Force (F) × Distance (d) × cos(θ)
Since the force is applied horizontally, the angle (θ) between the force and displacement is 0 degrees, and the cosine of 0 degrees is 1. So the equation becomes:
W = 25N × 3m × 1
W = 75 Joules
Therefore, the external work done on the block-table system is 75 Joules.
Step 2: Determine the energy dissipated by friction.
The energy dissipated by friction is equal to the work done against friction. Since the block is moving on the table, the frictional force opposes its motion. The equation to calculate the work done against friction is:
Work (W) = Force of friction (f) × Distance (d)
The force of friction can be calculated using the equation:
Force of friction (f) = coefficient of kinetic friction (µk) × Normal force (N)
The normal force can be determined as the weight of the block, which is equal to the mass (m) of the block multiplied by the acceleration due to gravity (g).
Normal force (N) = mass (m) × acceleration due to gravity (g)
Given that the coefficient of kinetic friction (µk) is 0.35 and the mass (m) of the block is 4kg, and the acceleration due to gravity (g) is 9.8m/s², we can calculate the normal force:
N = 4kg × 9.8m/s²
N = 39.2N
Next, we can calculate the force of friction:
f = µk × N
f = 0.35 × 39.2N
f = 13.72N
Finally, we can calculate the energy dissipated by friction:
W = f × d
W = 13.72N × 3m
W = 41.16 Joules
Therefore, the energy dissipated by friction is 41.16 Joules.
Step 3: Calculate the kinetic energy of the block.
The kinetic energy (KE) of an object is given by the formula:
KE = 0.5 × mass (m) × velocity²
Since the block is initially at rest, its initial velocity (u) is 0 m/s. The final velocity (v) can be determined using the equation:
v² = u² + 2 × acceleration (a) × distance (d)
Given that the force applied is horizontal, and no vertical forces act on the block, the acceleration is 0 m/s². Thus, the final velocity (v) is also 0 m/s, and the kinetic energy of the block is:
KE = 0.5 × 4kg × 0²
KE = 0 Joules
Therefore, the kinetic energy of the block is 0 Joules.
Step 4: Calculate the modulus of the velocity of the block.
The modulus (or magnitude) of the velocity (v) can be calculated using the equation:
Modulus of velocity = √(v_x² + v_y²)
In this case, since the block is only moving horizontally, its velocity in the y-direction is zero. Therefore, we only need to calculate the velocity in the x-direction, which is the final velocity (v).
v = u + acceleration × time
Since the block is moving at a constant speed, the velocity (v) is equal to the distance (d) divided by the time (t) taken to cover that distance.
v = d/t
v = 3m / t
However, we need to know the time (t) to calculate the velocity. Unfortunately, that information is not provided in the question. If you have the time, you can substitute it into the equation and calculate the modulus of the velocity of the block.
If you know the time, please provide it and I can help you calculate the modulus of the velocity.
To determine the external work on the block-table system, you need to calculate the work done by the applied force in pushing the block. The formula for work is given by:
Work = Force * Distance * cos(theta)
In this case, the force is 25 N (given), the distance is 3 m (given), and the angle (theta) between the force and direction of motion is 0 degrees since the force is horizontal. Therefore, cos(theta) = 1.
Substituting the values into the formula:
Work = 25 N * 3 m * 1 = 75 Joules
So, the external work on the block-table system is 75 Joules.
To find the energy dissipated by friction, you can use the formula:
Energy Dissipated = Force of Friction * Distance
The force of friction can be determined using the formula:
Force of Friction = coefficient of kinetic friction * Normal force
The Normal force can be calculated by:
Normal force = mass * gravity
In this case, the coefficient of kinetic friction is 0.35, the mass of the block is 4 kg, and the acceleration due to gravity is 9.8 m/s^2.
Normal force = 4 kg * 9.8 m/s^2 = 39.2 N
Force of Friction = 0.35 * 39.2 N = 13.72 N
Substituting the values into the formula to find energy dissipated:
Energy Dissipated = 13.72 N * 3 m = 41.16 Joules
So, the energy dissipated by friction is 41.16 Joules.
To calculate the kinetic energy of the block, you can use the formula:
Kinetic Energy = 0.5 * mass * velocity^2
Since the block was initially at rest, the velocity is the final velocity after being pushed.
To find the final velocity, you can use the equation:
Force - Force of Friction = mass * acceleration
In this case, the net force is the applied force, which is 25 N.
25 N - 13.72 N = 4 kg * acceleration
Acceleration = (25 N - 13.72 N) / 4 kg ≈ 2.07 m/s^2
Using the acceleration, you can find the final velocity:
v^2 - u^2 = 2as
v^2 - 0^2 = 2 * 2.07 m/s^2 * 3 m
v^2 = 12.42 m^2/s^2
Taking the square root:
v ≈ 3.52 m/s
Substituting the values into the kinetic energy formula:
Kinetic Energy = 0.5 * 4 kg * (3.52 m/s)^2 ≈ 24.64 Joules
So, the kinetic energy of the block is approximately 24.64 Joules.
To find the modulus of the velocity of the block, simply take the absolute value of the velocity, which is 3.52 m/s.
Therefore, the modulus of the velocity of the block is 3.52 m/s.