Question

What force is required to push a block (mass m) up an inclined plane that makes an angle of θ with the horizon at a constant velocity, if the coefficient of friction between the plane and the block is μ? Group of answer choices

a. mg μ cosθ
b. mg (μ cosθ + sinθ)
c. mg (μ sinθ + cosθ)
d. mg (μ cosθ + μ sinθ)
e. mg (μ cosθ + m sinθ)

Answer 1

The force needed to push a block up an incline at a constant velocity is the sum of the gravitational force parallel to the incline and the frictional force, which is mg sin(θ) + μmg cos(θ). So the correct option is e.

Step-by-step explanation:

The force required to push a block of mass m up an inclined plane at an angle θ with the horizon at a constant velocity, given the coefficient of friction between the plane and the block is μ, can be determined by analyzing the forces acting on the block. We need to consider both the component of the block's weight parallel to the incline and the frictional force.

Since the block is moving at a constant velocity, the net force along the incline must be zero. This implies that the applied force must counteract the combined forces of gravity pulling the block down the incline and the frictional force opposing the motion. The gravitational component parallel to the incline is mg sin(θ) and the frictional force is μmg cos(θ). Therefore, the required force F is:

F = mg sin(θ) + μmg cos(θ)

This corresponds to answer choice e.

Answer 2

b. mg ( μ · cos θ + sin θ)

Step-by-step explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following parallel forces acting on the block (in parallel direction to the direction of movement):

F = applied force.

Fr = friction force.

wx = parallel component of the weight.

According to Newton´s second law:

∑F = m · a

Where "m" is the mass of the block and "a" its acceleration.

Then:

F - Fr - wx = m · a

Since the block is to be pushed at a constant velocity, the acceleration is zero. Then:

F - Fr - wx = 0

F = Fr + wx

The applied force has to be equal to the friction force plus the parallel component of the weight to push the block at constant velocity.

The friction force is calculated as follows:

Fr = μ · N

Where N is the normal force and μ is the coefficient of friction.

Notice that the normal force is of the same magnitude as the perpendicular component of the weight, wy.

Let´s apply Newton´s second law in the perpendicular direction to show this:

∑F = m · a

N - wy = m · a

The acceleration of the block in the perpendicular direction is zero. Then:

N - wy = 0

N = wy

And wy can be obtained by trigonometry (see figure):

wy = W · cos θ

N = wy = mg · cos θ

The parallel component of the weight is calculated using trigonometry (see figure):

wx = W · sin θ

wx = mg · sin θ

Then the applied force will be:

F = Fr + wx

F = μ · N + mg · sin θ (N = wy = mg · cos θ)

F = μ · mg · cos θ + mg · sin θ

F = mg ( μ · cos θ + sin θ)

The correct answer is the b.