A 18 kg block is dragged over a rough, hor- izontal surface by a constant force of 131 N acting at an angle of 27 ◦ above the horizon- tal. The block is displaced 15 . 7 m, and the coefficient of kinetic friction is 0 . 125. The acceleration of gravity is 9 . 8 m / s 2 .If the block was originally at rest, determine its final speed. Answer in units of m / s.
Pulling Force= 131cos27
Friction Force= 0.125(18x9.8-131sin27)
Pulling Force-Friction Force= 18a
(solve for a)
Vf^2=Vi^2+2ad
(solve for Vf)
To determine the final speed of the block, we can use the work-energy principle.
The work done on an object is equal to the change in its kinetic energy. In this case, the work done can be broken down into two components: the work done by the applied force and the work done by friction.
The work done by the applied force can be calculated using the equation:
Work = Force * displacement * cos(theta),
where theta is the angle between the force and displacement vectors. In this case, theta is 27 degrees, so cos(theta) = cos(27) = 0.891.
Work_applied = Force * displacement * cos(theta)
= 131 N * 15.7 m * 0.891
= 2023.767 J
The work done by friction is given by:
Work_friction = Frictional force * displacement.
The frictional force can be calculated using the equation:
Frictional force = coefficient of friction * normal force,
where normal force = mass * acceleration due to gravity.
Normal force = mass * acceleration due to gravity
= 18 kg * 9.8 m/s^2
= 176.4 N
Frictional force = coefficient of friction * normal force
= 0.125 * 176.4 N
= 22.05 N
Work_friction = Frictional force * displacement
= 22.05 N * 15.7 m
= 346.185 J
The net work done on the block is the sum of the work done by the applied force and the work done by friction:
Net Work = Work_applied - Work_friction
= 2023.767 J - 346.185 J
= 1677.582 J
According to the work-energy principle, this net work is equal to the change in the kinetic energy of the block.
Kinetic energy at the end = Net Work
1/2 * mass * final velocity^2 = 1677.582 J
Rearranging the equation gives us:
final velocity^2 = 2 * Net Work / mass
final velocity^2 = 2 * 1677.582 J / 18 kg
final velocity^2 = 186.398 J/kg
Taking square root of both sides gives us:
final velocity = sqrt(186.398 J/kg)
final velocity = 13.641 m/s
Therefore, the final speed of the block is approximately 13.641 m/s.
To determine the final speed of the block, we need to break down the force acting on it and calculate the net force. Then we can use Newton's second law to find the acceleration of the block, and finally, use kinematic equations to find the final speed.
1. Break down the force: The applied force can be divided into two components: one parallel to the surface (horizontal direction) and one perpendicular to the surface (vertical direction).
- The horizontal component of the force is given by F_horizontal = F_applied * cos(theta), where F_applied is the applied force and theta is the angle it makes with the horizontal.
- The vertical component of the force is given by F_vertical = F_applied * sin(theta).
2. Calculate the force of friction: The force of friction opposes the motion and is given by the equation F_friction = coefficient of kinetic friction * normal force.
- The normal force is the force exerted by the surface on the block and is equal to the weight of the block, N = m * g, where m is the mass of the block and g is the acceleration due to gravity.
3. Determine the net force: The net force acting on the block is the vector sum of the applied force, the force of friction, and the gravitational force.
- The net force is given by F_net = F_horizontal - F_friction - F_gravity.
- The gravitational force, F_gravity, is equal to the weight of the block, which is F_gravity = m * g.
4. Calculate the acceleration: Using Newton's second law, F_net = m * a, we can determine the acceleration of the block.
- Rearranging the equation, a = F_net / m.
5. Use kinematic equations to find the final speed: Since the block starts from rest (initial velocity, vi = 0 m/s), we can use the equation v_f^2 = vi^2 + 2 * a * d, where v_f is the final velocity, a is the acceleration, and d is the displacement.
- Rearranging the equation, v_f = sqrt(2 * a * d).
Now, let's plug in the given values and calculate the final speed:
Given:
- Mass of the block (m) = 18 kg
- Applied force (F_applied) = 131 N
- Angle (theta) = 27 degrees
- Displacement (d) = 15.7 m
- Coefficient of kinetic friction (μ) = 0.125
- Acceleration due to gravity (g) = 9.8 m/s^2
Calculations:
1. Calculate the horizontal component of the applied force:
F_horizontal = F_applied * cos(theta)
= 131 N * cos(27 degrees)
≈ 115.163 N
2. Calculate the vertical component of the applied force:
F_vertical = F_applied * sin(theta)
= 131 N * sin(27 degrees)
≈ 58.442 N
3. Determine the force of friction:
F_friction = μ * N
= 0.125 * (m * g)
= 0.125 * (18 kg * 9.8 m/s^2)
≈ 21.735 N
4. Calculate the net force:
F_net = F_horizontal - F_friction - F_gravity
= F_horizontal - F_friction - (m * g)
= 115.163 N - 21.735 N - (18 kg * 9.8 m/s^2)
≈ 5.428 N
5. Calculate the acceleration:
a = F_net / m
= 5.428 N / 18 kg
≈ 0.301 m/s^2
6. Calculate the final speed:
v_f = √(2 * a * d)
= √(2 * 0.301 m/s^2 * 15.7 m)
≈ 3.42 m/s
Therefore, the final speed of the block is approximately 3.42 m/s.