A box of weight 2N accelerates down a rough plane that is inclined at an angle 30 degrees above the horizontal. The normal force acting on the box has a magnitude 1.7 N, the coefficient of kinetic friction between the box and the plane is O.30, and displacement of the box is 1.8 m down the inclined plane.
Two questions:
A. What is the work done on the box by the normal force?
B. What is the work done on the box by the force of kinetic friction?
Can someone give me the formulas to use?
The work done by normal force is zero, as work is force x distance in the direction of force.
frictionwork= .30*1.7*1.8 joules
A. The work done on the box by the normal force can be calculated using the formula:
Work = Force * Distance * cos(angle)
In this case, the normal force is acting perpendicular to the displacement of the box, so the angle between the normal force and the displacement is 90 degrees. Since cos(90 degrees) = 0, the work done by the normal force is 0.
B. The work done on the box by the force of kinetic friction can be calculated using the formula:
Work = Force * Distance * cos(angle)
The force of kinetic friction is given by:
Force of kinetic friction = Coefficient of kinetic friction * Normal force
Plugging the given values into the equation, we have:
Force of kinetic friction = 0.30 * 1.7 N = 0.51 N
Since the force of kinetic friction is acting opposite to the displacement, the angle between the force of kinetic friction and the displacement is 180 degrees. So, cos(180 degrees) = -1. Plugging in the values, we have:
Work = (-0.51 N) * 1.8 m * cos(180 degrees) = -0.51 N * 1.8 m * (-1) = 0.918 N · m or 0.918 Joules.
Therefore, the work done on the box by the force of kinetic friction is 0.918 Joules.
Sure! To find the work done on the box by the normal force, we can use the formula:
Work = Force * Distance * cos(angle)
Here, the force exerted by the normal force is perpendicular to the displacement of the box, so the angle between the force and the displacement is 90 degrees. Therefore, the cosine of 90 degrees is 0.
Therefore, the work done on the box by the normal force is 0, as there is no component of the normal force in the direction of motion.
To find the work done on the box by the force of kinetic friction, we can use the formula:
Work = Force * Distance * cos(angle)
Here, the force of kinetic friction can be calculated using the formula:
Force of kinetic friction = coefficient of kinetic friction * magnitude of the normal force
So, in this case, the force of kinetic friction can be calculated as:
Force of kinetic friction = 0.30 * 1.7 N
To find the work done, we need to know the displacement of the box along the inclined plane. Given that the displacement is 1.8 m down the inclined plane, we can use this value in the formula to find the work done on the box by the force of kinetic friction.
Therefore, the work done on the box by the force of kinetic friction is:
Work = (0.30 * 1.7 N) * (1.8 m) * cos(180 degrees - 30 degrees)
Using trigonometric identities, we can simplify the expression to:
Work = (0.30 * 1.7 N) * (1.8 m) * cos(150 degrees)
Now, use a calculator to find the value of cos(150 degrees), and substitute it into the formula, to get the final answer for the work done on the box by the force of kinetic friction.