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A block has a mass of 6kg and rest initially on a plane inclined at 35degrees. what value of force is required to stop the block from sliding if (a) the plane is frictionless, (b) the plane has coefficients of fractions static and kinetic equal to 0.20 and 0.18 respectively? what will the value of the force be if the velocity is constant and is directed upward parallel the incline?

User Knoobie
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Final answer:

To stop the block from sliding, the force required is equal to the weight of the block, regardless of the presence of friction. When the plane is frictionless, the force required is the weight of the block, which is 58.8 N. When the plane has coefficients of friction, the force of static friction is 9.8 N and the force of kinetic friction is 8.8 N.

Step-by-step explanation:

To calculate the force required to stop the block from sliding, we need to consider the forces acting on the block. When the plane is frictionless:

The force down the plane is equal to the weight of the block, which can be calculated as:

Weight = mass x gravity

where mass = 6 kg and gravity = 9.8 m/s² (acceleration due to gravity). Therefore, Weight = 6 kg x 9.8 m/s² = 58.8 N. As the block is at rest initially, there is no force opposing the motion down the plane since there is no friction. Thus, the force required to stop the block from sliding down the plane is equal to the weight of the block, which is 58.8 N.

When the plane has coefficients of friction static and kinetic equal to 0.20 and 0.18 respectively:

The force down the plane is still equal to the weight of the block, which is 58.8 N. However, there is now a force opposing the motion down the plane due to friction. The maximum force of static friction can be calculated as:

Force of static friction = coefficient of static friction x normal force

The normal force (N) can be calculated as:

Normal force = mass x gravity x cos(angle)

where angle = 35 degrees. Therefore, Normal force = 6 kg x 9.8 m/s² x cos(35 degrees) = 48.8 N.

Substituting the values into the equation, we have:

Force of static friction = 0.20 x 48.8 N = 9.8 N.

Therefore, the force required to stop the block from sliding when the plane has static friction is 9.8 N.

When the velocity is constant and directed upward parallel to the incline:

The force down the plane is still equal to the weight of the block, which is 58.8 N. But now, the force opposing the motion up the plane is equal to the force of kinetic friction, which can be calculated as:

Force of kinetic friction = coefficient of kinetic friction x normal force

Using the same normal force (48.8 N) and the coefficient of kinetic friction (0.18), we can calculate the force of kinetic friction as:

Force of kinetic friction = 0.18 x 48.8 N = 8.8 N.

Therefore, the force required to keep the block moving upward at a constant speed parallel to the incline is 8.8 N.

User Nadir SOUALEM
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