A 3.00 kg block starts from rest at the top of a 36.0'> incline and slides 2.00 m down the incline in 1.60 s.
(a) Find the acceleration of the block.
m/s2
(b) Find the coefficient of kinetic friction between the block and the incline.
(c) Find the frictional force acting on the block.
N
(d) Find the speed of the block after it has slid 2.00 m.
m/s
To find the answers to these questions, we can make use of several key principles of physics, such as Newton's laws of motion and equations related to motion on an inclined plane.
(a) First, let's find the acceleration of the block.
We can use the second equation of motion, which relates displacement (s), initial velocity (u), time (t), and acceleration (a):
s = ut + 0.5 * a * t^2
Given that the block starts from rest (u = 0) and slides down the incline for a distance of 2.00 m in 1.60 s, we can plug in these values into the equation:
2.00 m = 0 + 0.5 * a * (1.60 s)^2
Simplifying the equation, we get:
a = (2 * 2.00 m) / (1.60 s)^2
a = 2.50 m/s^2
So, the acceleration of the block is 2.50 m/s^2.
(b) To find the coefficient of kinetic friction between the block and the incline, we can use the equation that relates the net force acting on an object on an inclined plane to the frictional force:
Net force = m * g * sin(θ) - m * g * cos(θ) * μk
Where m is the mass of the block, g is the acceleration due to gravity, θ is the angle of the incline, and μk is the coefficient of kinetic friction.
In this case, we know the mass of the block is 3.00 kg, the angle of the incline is 36.0 degrees, and the acceleration due to gravity is approximately 9.8 m/s^2.
Let's calculate the net force acting on the block:
Net force = m * g * sin(θ) - m * g * cos(θ) * μk
0 = 3.00 kg * 9.8 m/s^2 * sin(36.0°) - 3.00 kg * 9.8 m/s^2 * cos(36.0°) * μk
Simplifying the equation, we get:
μk = sin(36.0°) / cos(36.0°)
μk ≈ 0.587
So, the coefficient of kinetic friction between the block and the incline is approximately 0.587.
(c) To find the frictional force acting on the block, we can use the equation for frictional force:
Frictional force = m * g * cos(θ) * μk
Plugging in the given values, we get:
Frictional force = 3.00 kg * 9.8 m/s^2 * cos(36.0°) * 0.587
Frictional force ≈ 16.14 N
So, the frictional force acting on the block is approximately 16.14 N.
(d) Finally, to find the speed of the block after it has slid 2.00 m, we can use the third equation of motion, which relates final velocity (v), initial velocity (u), acceleration (a), and displacement (s):
v^2 = u^2 + 2 * a * s
Given that the block starts from rest (u = 0), we can plug in the values into the equation:
v^2 = 0 + 2 * 2.50 m/s^2 * 2.00 m
Simplifying the equation, we get:
v^2 = 10.0 m^2/s^2
Taking the square root of both sides, we obtain:
v ≈ 3.16 m/s
So, the speed of the block after it has slid 2.00 m is approximately 3.16 m/s.