A 2.84 kg block starts from rest at the top of a 30° incline and accelerates uniformly down the incline, moving 1.92 m in 1.90 s.
(a) Find the magnitude of the acceleration of the block.
m/s2
(b) Find the coefficient of kinetic friction between the block and the incline.
(c) Find the magnitude of the frictional force acting on the block.
N
(d) Find the speed of the block after it has slid a distance 1.92 m.
m/s
a. d = 0.5a*t^2 = 1.92 m.
0.5a*1.9^2 = 1.92 m
1.81a = 1.92
a = 1.06 m/s^2
b. M*g = 2.84 * 9.8 = 27.8 N. = Wt. of
block.
Fp = 27.8*sin30 = 13.9 N. = Force parallel to the incline.
Fn = 27.8*Cos30 = 24.1 N. = Normal force
Fp-Fk = m*a
13.9-Fk = 2.84*1.06 = 3.01
-Fk = 3.01-13.9 = -10.9
Fk = 10.9 N. = Force of kinetic energy.
uk = Fk/Fn = 10.9/24.1 = 0.452
c. Fk = 10.9 N. See part b.
d. V^2 = Vo^2 + 2a*d
Vo = 0
a = 1.06 m/s^2
d = 1.92 m.
Solve for V.
(a) To find the magnitude of the acceleration of the block, we can use the kinematic equation:
\[s = ut + \frac{1}{2}at^2\]
Where:
- s is the distance traveled (1.92 m),
- u is the initial velocity (0 m/s since the block starts from rest),
- t is the time taken (1.90 s), and
- a is the acceleration that we need to find.
Rearranging the equation to solve for a, we get:
\[a = \frac{2\left(s - ut\right)}{t^2}\]
Substituting the given values, we have:
\[a = \frac{2\left(1.92 - (0)(1.90)\right)}{(1.90)^2}\]
Calculating this expression will give us the magnitude of the acceleration of the block in m/s².
(b) To find the coefficient of kinetic friction between the block and the incline, we need to identify the forces acting on the block. In this case, the gravitational force can be broken down into two components: the force parallel to the incline (Fₚ) and the force perpendicular to the incline (Fₑ). The force parallel to the incline is counteracted by the frictional force (Fₖ). The equation relating these forces is given by:
\[ Fₑ = mg \cos θ \]
\[ Fₚ = mg \sin θ \]
\[ Fₖ = μ_kFₑ \]
Where:
- m is the mass of the block (2.84 kg),
- g is the acceleration due to gravity (9.8 m/s²),
- θ is the angle of the incline (30°),
- μ_k is the coefficient of kinetic friction that we need to find.
Substituting the given values into the equations, we can solve for the coefficient of kinetic friction (μ_k).
(c) To find the magnitude of the frictional force acting on the block, we can use the equation mentioned before:
\[ Fₖ = μ_kFₑ \]
Where Fₖ is the frictional force and Fₑ is the force perpendicular to the incline.
Substitute the known values and solve for Fₖ.
(d) The speed of the block after it has slid a distance of 1.92 m can be found using the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- v is the final velocity that we need to find,
- u is the initial velocity (0 m/s since the block starts from rest),
- a is the acceleration (which we calculated in part (a)),
- s is the distance traveled (1.92 m).
Rearranging the equation to solve for v, we get:
\[ v = \sqrt{u^2 + 2as} \]
Substituting the given values into the equation will give us the speed of the block after it has slid a distance of 1.92 m.