A crate of mass 29 kg is pushed up a ramp by a person as shown in the figure below. The person pushes the crate a distance of 25 m as measured along the ramp. Assume the crate moves at constant velocity. Assume the coefficient of kinetic friction between the crate and the ramp is
μK = 0.22.
What is the work done by the person as he pushes the crate up the ramp?
W=ΔPE+W(fr)=
=m•g•Δh+μ•m•g•cosα•s=
=m•g•s•sinα+μ•m•g•cosα•s=
=m•g•s(sinα + μ•cosα)
The work done by the person can be calculated using the formula:
Work = Force * Distance * cos(theta)
In this case, the force is the force applied by the person, the distance is the distance the crate is moved along the ramp, and theta is the angle between the force and the displacement of the crate.
Since the crate is moving at a constant velocity, the net force acting on it must be zero. This means that the force applied by the person must be equal and opposite to the force of kinetic friction acting on the crate.
The force of kinetic friction, Fk, can be calculated using the formula:
Fk = μk * Fn
where μk is the coefficient of kinetic friction and Fn is the normal force. The normal force, Fn, is equal to the weight of the crate, which can be calculated using the formula:
Fn = m * g
where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now we can calculate the work done by the person:
Work = Force * Distance * cos(theta)
Since the force applied by the person is equal and opposite to the force of kinetic friction, the work done by the person can be calculated using the formula:
Work = Fk * Distance * cos(theta)
Substituting the values we have:
Work = (μk * Fn) * Distance * cos(theta)
Work = (μk * m * g) * Distance * cos(theta)
Now we can plug in the given values:
m = 29 kg
μk = 0.22
Distance = 25 m
g = 9.8 m/s^2
Work = (0.22 * 29 kg * 9.8 m/s^2) * 25 m * cos(theta)
Since the angle theta is not mentioned in the question, we cannot determine the exact value of cos(theta) without additional information.
To find the work done by the person as they push the crate up the ramp, we need to consider the force exerted by the person and the distance over which the force is applied.
The work done is given by the formula:
Work = Force × Distance × cos(θ)
Where:
Force is the component of the force parallel to the direction of motion,
Distance is the distance over which the force is applied,
and θ is the angle between the force and the direction of motion.
In this case, the force exerted by the person is the force required to overcome the force of friction and the force due to gravity acting on the crate.
The force required to overcome friction can be found using the formula:
Friction force = coefficient of kinetic friction × normal force
The normal force is the force exerted by the ramp on the crate, which is equal in magnitude and opposite in direction to the force due to gravity. Therefore, the normal force can be found using the formula:
Normal force = mass × gravitational acceleration
Next, we need to find the component of the force exerted by the person that is parallel to the direction of motion. This can be found using simple trigonometry:
Force parallel = Force × sin(θ)
Since the crate is moving at constant velocity, the force exerted by the person must be equal in magnitude and opposite in direction to the net force acting on the crate. Therefore, we have:
Force parallel = Friction force + Force due to gravity
Now, we can calculate the work done. We have all the required values:
Work = Force parallel × Distance × cos(θ)
Substituting the values, we can find the work done by the person.