Starting from rest, a 5kg block slides 2.5 m down a rough 30o incline in 2 seconds. Find
a) the work done by the force of gravity.
b) the work done by friction.
c) the work done by the normal force
Do part b last
To find the work done by different forces on the block, we need to consider the work-energy principle. The work done by a force is equal to the force applied multiplied by the displacement over which the force is applied.
First, let's find the gravitational force acting on the block. The gravitational force is given by the equation:
F_gravity = m * g
where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the mass of the block is 5 kg, so the gravitational force is:
F_gravity = 5 kg * 9.8 m/s^2 = 49 N
Now, let's find the work done by the force of gravity. Since the block slides down the incline, the force of gravity acts in the same direction as the displacement. Thus, the work done by the force of gravity is:
Work_gravity = F_gravity * d * cos(theta)
where d is the displacement (2.5 m) and theta is the angle of the incline (30 degrees).
Work_gravity = 49 N * 2.5 m * cos(30 degrees) = 106.23 J
Therefore, the work done by the force of gravity is 106.23 Joules.
Now, let's move on to finding the work done by friction. We'll use the equation:
Work_friction = force_friction * d * cos(theta)
To find the force of friction, we need to use the equation:
force_friction = coefficient_friction * force_normal
The force normal (force perpendicular to the incline) can be found using the equation:
force_normal = m * g * cos(theta)
where theta is the angle of the incline.
force_normal = 5 kg * 9.8 m/s^2 * cos(30 degrees) = 42.45 N
Now, we need the coefficient of friction. Let's assume it is given as μ.
Given that we need to calculate the work done by friction last (part b), we'll skip the calculation for now and move on to finding the work done by the normal force (part c).
The normal force is the perpendicular force exerted by a surface on an object. In this case, it's exerted by the incline. It can be calculated using the equation:
force_normal = m * g * cos(theta)
Here, the force normal is also equal to the force exerted by the incline on the block.
force_normal = 5 kg * 9.8 m/s^2 * cos(30 degrees) = 42.45 N
The work done by the normal force is zero since it acts perpendicular to the displacement.
To calculate the work done by friction (part b), we'll use the equation:
Work_friction = force_friction * d * cos(theta)
So now, we need to calculate the force_of_friction. Rearranging the formula above, we get:
force_friction = Work_friction / (d * cos(theta))
However, to find the work done by friction, we need the coefficient of friction (μ). Assuming it is given, we can use the equation:
force_friction = coefficient_friction * force_normal
Substituting the given values, we get:
force_friction = μ * 42.45 N
Finally, substitute the value of force_friction back into the work equation:
Work_friction = force_friction * d * cos(theta)
This will give us the work done by friction (part b).