You push a block of mass m up a plane inclined at angle θ
at constant velocity v with a force F parallel to the inclined plane. The coefficient of kinetic friction between the block and the plane is µk. Which is the correct equation for F? (g = gravitational acceleration)
A. F = mg
B. F = mg ( sin θ − µk cos θ ) C. F = mg ( sin θ + µk cos θ ) D. F = µkmg ( cos θ )
A car of mass m travels at constant speed v around a traffic circle of radius r. If the driver wanted to push the limits and drive around the traffic circle at twice the speed, determine the force F that would be necessary to keep the car traveling around the traffic circle at constant speed 2v?
A. F = 4mv^2/r r^
B. F = 2mv^2/r r^
C. F = -4mv^2/r r^
D. F = - 2mv^2/r r^
and what are your thoughts?
For the first question:
The correct equation for F is B. F = mg ( sin θ − µk cos θ )
Explanation:
When the block is moving at constant velocity up the inclined plane, the force F applied parallel to the plane must balance out the gravitational force pulling the block down the plane and the force of kinetic friction opposing the motion.
The gravitational force pulling the block down the plane is mg sin θ (where θ is the angle of the inclined plane).
The force of kinetic friction opposing the motion is µk mg cos θ (where µk is the coefficient of kinetic friction and mg is the normal force).
Therefore, the net force (F) must be equal to the sum of these two forces:
F = mg sin θ - µk mg cos θ.
Simplifying this equation gives:
F = mg ( sin θ − µk cos θ )
To solve these physics problems, we need to analyze the forces acting on the objects and make use of relevant equations.
1) In the first problem, a block is being pushed up an inclined plane with constant velocity. The forces acting on the block are the force of gravity (mg), the normal force (N), and the force of kinetic friction (fk). Since the block is moving at constant velocity, the net force acting on it is zero.
The force applied parallel to the inclined plane (F) is countered by the force of kinetic friction. The magnitude of the force of kinetic friction is given by fk = µkN, where µk is the coefficient of kinetic friction.
Now, let's break down the forces involved. The force of gravity can be split into two components: mg sinθ acting parallel to the incline and mg cosθ acting perpendicular to the incline.
The normal force, N, acts perpendicular to the incline. It cancels out the mg cosθ component of the force of gravity.
Since the block is moving at constant velocity, the magnitude of the force applied parallel to the incline (F) must be equal to the magnitude of the force of kinetic friction (fk): F = fk.
Substituting fk = µkN, we can express N as mg cosθ: N = mg cosθ.
Substituting N back into the equation, we have F = µk(mg cosθ).
The correct equation for F is option D: F = µkmg cosθ.
2) In the second problem, a car is traveling at constant speed v around a traffic circle of radius r. We need to determine the force F required to keep the car traveling at twice the speed (2v).
In circular motion, the centripetal force (F) required to keep an object moving in a circle is given by F = mv^2/r, where m is the mass of the object, v is the speed, and r is the radius of the circle.
In this case, the mass of the car remains the same, but the speed changes from v to 2v. Substituting these values into the formula, we get F = (m(2v)^2)/r.
Simplifying, we have F = 4mv^2/r.
Therefore, the correct equation for F is option A: F = 4mv^2/r.