A 1.20*10^2 kg crate is being pushed across a horizontal floor by a force P that makes an angle of 30.0° below the horizontal. The coefficient of kinetic friction is 0.350. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?
To find the magnitude of force P, we need to consider the net work done by the force P and the kinetic frictional force.
1. The net work done by force P can be calculated using the formula: Net work = force x displacement x cosine(theta). We want the net work to be zero, so:
Net work = P x displacement x cosine(30°) = 0
2. The displacement is the distance the crate is pushed across the floor. This information is not provided in the question, so we'll assume it to be d.
P x d x cosine(30°) = 0
3. The kinetic frictional force can be calculated using the equation: kinetic frictional force = μ * normal force. The normal force is equal to the weight of the crate, which is given by: weight = mass x gravitational acceleration.
kinetic frictional force = μ * (mass x gravitational acceleration)
4. Substituting the given values, we have:
kinetic frictional force = 0.350 x (1.20 x 10^2 kg x 9.8 m/s^2)
5. Since the net work done by the force P and the kinetic frictional force is zero, the magnitude of force P is equal to the kinetic frictional force.
P = kinetic frictional force
So, the magnitude of force P should be equal to the calculated kinetic frictional force value.
To find the magnitude of force P, we need to determine the net work done by it and the kinetic frictional force.
Net work is defined as the total work done by all forces acting on an object. In this case, the net work done should be zero.
Let's break down the forces acting on the crate:
1. Force P: This is the applied force that pushes the crate across the floor. Its magnitude is what we need to find.
2. Normal force (Fn): This is the force exerted by the floor on the crate perpendicular to the surface. It is equal in magnitude and opposite in direction to the weight of the crate. Fn = mg, where m is the mass of the crate and g is the acceleration due to gravity (9.8 m/s^2).
3. Frictional force (Ff): This is the force opposing the motion of the crate. It is equal to the coefficient of kinetic friction (μk) multiplied by the normal force. Ff = μk * Fn.
The net work is given by the equation:
Net work = work done by force P - work done by frictional force
Work done by force P is the product of the magnitude of force P, the displacement of the crate, and the cosine of the angle between the force and displacement.
Work done by frictional force is the product of the magnitude of frictional force, the displacement of the crate, and the cosine of the angle between the force and displacement (which is 180 degrees in this case, because friction is opposite to the direction of motion).
Since the net work should be zero, we can set up the equation:
0 = (P * d * cosθ) - (Ff * d * cos180°)
Now let's substitute the values given in the question:
m = 1.20 * 10^2 kg
θ = 30° below the horizontal
μk = 0.350
Fn = mg = (1.20 * 10^2 kg) * (9.8 m/s^2) = 1.18 * 10^3 N
Ff = μk * Fn = (0.350) * (1.18 * 10^3 N) = 4.13 * 10^2 N
Since cos180° = -1, the equation simplifies to:
0 = P * d * cosθ + Ff * d
Now we have an equation with one unknown (P). We can solve for P:
P * d * cosθ = - Ff * d
P = (- Ff * d) / (d * cosθ)
Simplifying:
P = - Ff / cosθ
P = - (4.13 * 10^2 N) / cos30°
P = - (4.13 * 10^2 N) / (√3 / 2)
P = - (4.13 * 10^2 N) * (2 / √3)
P ≈ - 4.77 * 10^2 N
Since force is a vector quantity, the negative sign indicates that the force P should be in the opposite direction of the frictional force. Therefore, the magnitude of P should be approximately 4.77 * 10^2 N.