A 3.00-kg block starts from rest at the top of a 33.5° incline and slides 2.00 m down the incline in 1.30 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
See 11-22-11,6:40pm post.
To find the answers to the given questions, we can use the principles of Newtonian mechanics and the equations of motion. Let's break down each question and go step by step to find the required values.
(a) Find the acceleration of the block:
To calculate the acceleration of the block, we can use the equation of motion:
d = v0t + (1/2)at^2
where
d = displacement (2.00 m)
v0 = initial velocity (0 m/s)
t = time (1.30 s)
a = acceleration
Rearranging the equation, we get:
a = (2d - v0t) / t^2
Plugging in the values, we have:
a = (2 * 2.00 m - 0 m/s * 1.30 s) / (1.30 s)^2
Solving for a, we find:
a = 1.21 m/s^2
Therefore, the acceleration of the block is 1.21 m/s^2.
(b) Find the coefficient of kinetic friction between the block and the incline:
To find the coefficient of kinetic friction, we need to use the relationship between the frictional force, the normal force, and the coefficient of friction:
f_friction = coefficient_friction * f_normal
Since the block is on an inclined plane, we need to split the gravitational force into its components parallel and perpendicular to the incline:
f_parallel = m * g * sin(theta)
f_perpendicular = m * g * cos(theta)
where
m = mass of the block (3.00 kg)
g = acceleration due to gravity (9.8 m/s^2)
theta = angle of the incline (33.5°)
The normal force f_normal is equal but opposite to the perpendicular force:
f_normal = f_perpendicular
Using Newton's second law, we can relate the net force acting parallel to the incline with the mass and acceleration (assuming there are no other horizontal forces):
f_parallel - f_friction = m * a
Substituting the relevant equations, we have:
m * g * sin(theta) - coefficient_friction * m * g * cos(theta) = m * a
Simplifying and solving for the coefficient of friction:
coefficient_friction = (m * g * sin(theta) - m * a) / (m * g * cos(theta))
Plugging in the values, we get:
coefficient_friction = (3.00 kg * 9.8 m/s^2 * sin(33.5°) - 3.00 kg * 1.21 m/s^2) / (3.00 kg * 9.8 m/s^2 * cos(33.5°))
Evaluating this expression gives us:
coefficient_friction ≈ 0.238
Therefore, the coefficient of kinetic friction between the block and the incline is approximately 0.238.
(c) Find the frictional force acting on the block:
To find the frictional force, we can use the equation:
f_friction = coefficient_friction * f_normal
Using the calculated value of the coefficient of friction from part (b), we have:
f_friction = 0.238 * f_normal
Now, we can substitute the expression for f_normal in terms of the mass and gravitational force:
f_friction = 0.238 * (m * g * cos(theta))
Plugging in the known values, we get:
f_friction = 0.238 * (3.00 kg * 9.8 m/s^2 * cos(33.5°))
Calculating this expression gives us:
f_friction ≈ 8.84 N
Therefore, the frictional force acting on the block is approximately 8.84 N.
(d) Find the speed of the block after it has slid 2.00 m:
To find the final speed of the block, we can use the equation of motion:
v^2 = v0^2 + 2ad
where
v = final velocity (unknown)
v0 = initial velocity (0 m/s)
a = acceleration (1.21 m/s^2)
d = displacement (2.00 m)
Rearranging the equation, we have:
v = sqrt(v0^2 + 2ad)
Since the initial velocity is zero, the equation simplifies to:
v = sqrt(2ad)
Plugging in the values, we get:
v = sqrt(2 * 1.21 m/s^2 * 2.00 m)
Calculating this expression gives us:
v ≈ 2.20 m/s
Therefore, the speed of the block after it has slid 2.00 m is approximately 2.20 m/s.