In the figure, a 6-kg box is sliding up an incline that makes an angle of theta=30 degree (with respect to the horizontal). As it slides up the incline, a kinetic frictional force fk acts on it. The acceleration of the block has magnitude 8 m/s2. Let the x+ direction point up the plane and the y+ direction be perpendicular to the plane.
(A) draw free body diagram **I did this
(B) What of the following statements correctly describes the x component ax of acceleration and the y component ay of acceleration as the block slides up the inclined plane? **answer: ax is negative, while ay is zero
(C) What is the normal force acting on the box as it slides up the inclined plane? **answer: FN = 51.96 N
(D) What is the kinetic frictional force acting on the box as it slides up the inclined plane? **answer: 18 N
I do not need the answers. I only need help understanding why the answer is the answer for part (B) and help finding out how the answer was gotten for part (D).
Sure! Let's break down each part of the question and explain how to arrive at the answer.
(B) The x component (ax) of acceleration refers to the acceleration in the direction parallel to the incline, and the y component (ay) refers to the acceleration in the direction perpendicular to the incline.
In this case, the block is sliding up the inclined plane, which means it is moving vertically. Therefore, the y component (ay) of acceleration is zero since there is no vertical acceleration. Gravity is acting vertically downward, but the normal force from the incline counteracts it, resulting in zero acceleration in the y direction.
The x component (ax) of acceleration is negative because the block is sliding up the incline against the force of kinetic friction (fk). The inclined plane acts as a ramp, and since the acceleration opposes the direction of motion, it has a negative value.
(C) To find the normal force (FN) acting on the box, we need to consider the forces acting on the block in the y direction. The normal force acts perpendicular to the incline and balances the component of gravity acting perpendicular to the incline.
The formula to calculate the normal force is FN = mg * cos(theta), where m is the mass of the box (6 kg) and theta is the angle of the incline (30 degrees).
FN = 6 kg * 9.8 m/s^2 * cos(30 degrees)
= 6 kg * 9.8 m/s^2 * (sqrt(3)/2)
≈ 51.96 N
(D) The kinetic frictional force (fk) acting on the box as it slides up the inclined plane can be found using the formula fk = uk * FN, where uk is the coefficient of kinetic friction.
To calculate fk, we need to know the coefficient of kinetic friction. If it is not given in the question, we won't be able to determine the exact value. However, if you have been provided with the coefficient of kinetic friction, you can simply multiply it by the normal force (FN) calculated in part (C).
For example, if the coefficient of kinetic friction is 0.3,
fk = 0.3 * 51.96 N
≈ 15.59 N
Please note that the actual value of the kinetic frictional force depends on the coefficient of kinetic friction, which determines the roughness between the contact surfaces.