A 10 kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20 degrees with the horizontal. If the coefficient of kinetic friction is .40 and the crate is pulled a distance of 5 m, (a) how much work is done by gravitational force? (b) How much work is done by the 100 N force? (c) What is the change in kinetic energy of the crate? (d) What is the speed of the crate after it is pulled 5m?
Before you start answering the various parts of the problem, it is a good idea to draw a free body diagram of forces acting on the block.
The force pulling it up the ramp is
F=100 N
The gravity force (weight) downwards is
M g = 10*9.8 = W = 98 N
The normal force appled to the block by the incline is
Fn = M g cos 20 = 92.1 N
The friction force is
Ff = Fn*Uk = 36.8 N (backwards)
Vertical height change
H = 5 sin 20 = 1.71 m
(a) - W*H = ? (it is negative because the object is raised against the gravity force
(b) F * 5m = ?
(c) (F - Ff)* 5 - W*H = ?
(d) =The answer to (c) equals the kinetic energy change. Use that to get the final kinetic energy. That can tell you what the final speed is, using the rule that K.E. = (1/2) M V^2
To solve this problem, we will need to use the following formulas:
(a) The work done by gravitational force is given by the equation: Work = Force x Distance x cos(angle)
(b) The work done by the 100 N force is given by the equation: Work = Force x Distance x cos(angle)
(c) The change in kinetic energy is given by the equation: Change in Kinetic Energy = Work done by the 100 N force - Work done by friction
(d) The speed of the crate after being pulled is given by the equation: Final speed = sqrt((initial speed)^2 + (2 x acceleration x distance))
Now, let's calculate each step:
(a) To calculate the work done by gravitational force:
The force of gravity on the crate is given by: Force of gravity = mass x gravitational acceleration
Force of gravity = 10 kg x 9.8 m/s^2 (assuming downward direction) = 98 N
The angle between the force of gravity and the direction of motion is 90 degrees, so the cosine of 90 degrees is 0.
Therefore, Work (gravitational) = 98 N x 5 m x cos(90 degrees) = 0 J (Joules)
(b) To calculate the work done by the 100 N force:
Work (100 N force) = 100 N x 5 m x cos(20 degrees) (given angle in the question) = 452.27 J (Joules)
(c) To calculate the change in kinetic energy:
First, we need to calculate the work done by friction.
The normal force (perpendicular to the incline) is given by: Normal force = mass x gravitational acceleration x cos(angle)
Normal force = 10 kg x 9.8 m/s^2 x cos(20 degrees) = 91.24 N
The force of friction is given by: Force of friction = coefficient of kinetic friction x normal force
Force of friction = 0.40 x 91.24 N = 36.50 N
Work (friction) = Force of friction x Distance x cos(180 degrees) (since friction acts opposite to the motion)
Work (friction) = 36.50 N x 5 m x cos(180 degrees) = -182.50 J (Joules)
Change in kinetic energy = Work (100 N force) - Work (friction)
Change in kinetic energy = 452.27 J - (-182.50 J) = 634.77 J (Joules)
(d) To calculate the final speed of the crate:
The force parallel to the incline is given by: Force parallel = 100 N x sin(20 degrees)
Force parallel = 34.20 N
The net force acting on the crate is given by: Net force = Force parallel - Force of friction
Net force = 34.20 N - 36.50 N = -2.30 N (since the net force acts opposite to the motion)
We can use Newton's second law to find the acceleration of the crate.
Acceleration = Net force / mass = (-2.30 N) / 10 kg = -0.23 m/s^2 (assuming the downward direction is positive)
Finally, we can use the equation for final speed to find the speed of the crate after being pulled 5 m:
Final speed = sqrt((1.5 m/s)^2 + (2 x -0.23 m/s^2 x 5 m)) = 1.21 m/s (rounded to two decimal places)
Therefore, the answers are:
(a) The work done by gravitational force is 0 J.
(b) The work done by the 100 N force is 452.27 J (rounded to two decimal places).
(c) The change in kinetic energy of the crate is 634.77 J (rounded to two decimal places).
(d) The speed of the crate after being pulled 5 m is 1.21 m/s (rounded to two decimal places).
To answer these questions, we need to understand the concepts of work, gravitational force, kinetic friction, and change in kinetic energy. Let's break down the problem step by step.
(a) To calculate the work done by the gravitational force, we need to find the component of the weight force parallel to the incline. We can do this using the formula:
Force_parallel = Weight * sin(angle)
Weight = mass * gravitational acceleration
Given that the mass of the crate is 10 kg and the gravitational acceleration is approximately 9.8 m/s^2, we can calculate the weight:
Weight = 10 kg * 9.8 m/s^2 = 98 N
Now, we can find the component of weight parallel to the incline:
Force_parallel = 98 N * sin(20 degrees) = 33.460 N
The work done by the gravitational force can be calculated using the formula:
Work = Force_parallel * distance * cos(angle)
Here, the angle is the angle of the incline. Since it is not given, we cannot calculate the exact value without further information. Let's assume the angle is 30 degrees for this example.
Work = 33.460 N * 5 m * cos(30 degrees) = 145.439 J
Therefore, the work done by the gravitational force is approximately 145.439 J.
(b) The work done by the 100 N force can be calculated using the formula:
Work = Force * distance * cos(angle)
Given that the force is 100 N and the angle is 20 degrees, and the distance is 5 m, we can calculate the work:
Work = 100 N * 5 m * cos(20 degrees) = 471.604 J
Therefore, the work done by the 100 N force is approximately 471.604 J.
(c) The change in kinetic energy of the crate can be calculated using the formula:
Change in Kinetic Energy = Work done by the net force
The net force can be calculated by subtracting the force of kinetic friction from the force applied. The force of kinetic friction can be found using the formula:
Force_friction = coefficient of kinetic friction * normal force
Normal force = Weight * cos(angle)
Given that the coefficient of kinetic friction is 0.40, we can calculate the force of kinetic friction:
Normal force = 98 N * cos(20 degrees) = 92.786 N
Force_friction = 0.40 * 92.786 N = 37.114 N
Now we can find the net force:
Net force = Force_applied - Force_friction
Force_applied = 100 N (given)
Net force = 100 N - 37.114 N = 62.886 N
Using the work formula, we can calculate the change in kinetic energy:
Change in Kinetic Energy = Work = Net force * distance * cos(angle)
Given that the distance is 5 m and the angle is 20 degrees, we have:
Change in Kinetic Energy = 62.886 N * 5 m * cos(20 degrees) = 535.834 J
Therefore, the change in kinetic energy of the crate is approximately 535.834 J.
(d) To find the speed of the crate after it is pulled 5 m, we can use the principle of conservation of energy. The initial kinetic energy of the crate is 0.5 * mass * velocity^2, and the final kinetic energy is given by the change in kinetic energy calculated in (c).
0.5 * mass * initial velocity^2 = final kinetic energy
Given that the mass is 10 kg and the initial velocity is 1.5 m/s, we can solve for the final velocity:
0.5 * 10 kg * (1.5 m/s)^2 = final kinetic energy
Final kinetic energy = 535.834 J (from part c)
Rearranging the equation:
final kinetic energy = 0.5 * 10 kg * final velocity^2
Solving for the final velocity:
final velocity^2 = (2 * final kinetic energy) / (10 kg)
final velocity^2 = (2 * 535.834 J) / (10 kg)
final velocity^2 = 535.834 J / 5 kg
final velocity^2 = 107.1668 m^2/s^2
Taking the square root:
final velocity ā ā(107.1668 m^2/s^2) ā 10.35 m/s
Therefore, the speed of the crate after it is pulled 5 m is approximately 10.35 m/s.