A 10 kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. The pulling force is 86 N parallel to the incline, which makes an angle of 30 ° with the horizontal. Assuming the coefficient of kinetic friction is 0.4 and the crate is pulled a distance of 7.5 mfind the following:
a) the work done by the Earth's gravity on the crate
b) the work done by the force of friction on the crate
c) the work done by the puller on the crate
d) the change in kinetic energy of the crage
e) the speed of the crate after it is pulled the 7.5 m
To answer these questions, we need to use the concepts of work and energy. Let's go through each question step by step:
a) The work done by the Earth's gravity on the crate:
The work done by gravity can be calculated using the formula:
Work = Force * Distance * cos(θ)
where Force is the weight of the crate (mg) and θ is the angle between the force and the direction of motion. In this case, the crate is moving up the incline, so the angle between the force of gravity and the direction of motion is 180 °.
The weight of the crate can be calculated by multiplying its mass (10 kg) by the acceleration due to gravity (9.8 m/s^2). Thus, the force due to gravity is:
Force = 10 kg * 9.8 m/s^2 = 98 N
Since the angle is 180 °, cos(180 °) = -1. Therefore, the work done by gravity is:
Work = -98 N * 7.5 m * cos(180 °) = -735 J
The negative sign indicates that the work done by gravity is against the motion of the crate.
b) The work done by the force of friction on the crate:
The work done by friction can be calculated using the formula:
Work = Force * Distance * cos(θ)
where Force is the force of friction, Distance is the distance the crate is pulled, and θ is the angle between the force of friction and the direction of motion. In this case, the crate is moving up the incline, so the angle between the force of friction and the direction of motion is 180 °.
The force of friction can be calculated using the formula:
Force of friction = coefficient of kinetic friction * Normal force
The Normal force is equal to the weight of the crate (98 N) times the cosine of the angle between the incline and the vertical direction (θ = 30 °). Therefore,
Normal force = 98 N * cos(30 °) = 85 N
Using the given coefficient of kinetic friction (0.4), the force of friction is:
Force of friction = 0.4 * 85 N = 34 N
Since the angle is 180 °, cos(180 °) = -1. Therefore, the work done by friction is:
Work = -34 N * 7.5 m * cos(180 °) = -255 J
The negative sign indicates that the work done by friction is against the motion of the crate.
c) The work done by the puller on the crate:
The work done by the puller can be calculated using the formula:
Work = Force * Distance * cos(θ)
where Force is the pulling force (86 N), Distance is the distance the crate is pulled, and θ is the angle between the pulling force and the direction of motion. In this case, the angle is 30 °.
Since the angle is 30 °, cos(30 °) = √3/2. Therefore, the work done by the puller is:
Work = 86 N * 7.5 m * cos(30 °) = 448 J
d) The change in kinetic energy of the crate:
The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. Therefore, the change in kinetic energy of the crate is equal to the sum of the work done by gravity, the work done by friction, and the work done by the puller:
Change in kinetic energy = Work by gravity + Work by friction + Work by puller
Change in kinetic energy = -735 J + (-255 J) + 448 J = -542 J
The negative sign indicates a decrease in kinetic energy.
e) The speed of the crate after it is pulled the 7.5 m:
To find the final speed of the crate, we can use the principle of conservation of energy. The initial kinetic energy of the crate is:
Initial kinetic energy = 0.5 * mass * (initial speed)^2
Initial kinetic energy = 0.5 * 10 kg * (1.5 m/s)^2 = 11.25 J
Since the work done on the crate results in a decrease in kinetic energy, the final kinetic energy is:
Final kinetic energy = Initial kinetic energy + Change in kinetic energy
Final kinetic energy = 11.25 J - 542 J = -530.75 J
The final kinetic energy is negative because the crate comes to a stop.
The final speed of the crate can be calculated by rearranging the formula for kinetic energy:
Final kinetic energy = 0.5 * mass * (final speed)^2
-530.75 J = 0.5 * 10 kg * (final speed)^2
Solving for the final speed:
(final speed)^2 = (-530.75 J) / (0.5 * 10 kg) = -53.075 m^2/s^2
Since speed cannot be negative, we ignore the negative sign and take the square root:
final speed = √(53.075 m^2/s^2) = 7.28 m/s (rounded to two decimal places)
Therefore, the speed of the crate after it is pulled the 7.5 m is approximately 7.28 m/s.