A 2.5-kg block slides down a 17.0° inclined plane with constant acceleration. The block starts from rest at the top. At the bottom its velocity reaches 0.70 m/s. The length of the incline is 1.7 m.
What is the acceleration of the block?
What is the coefficient of friction between the plane and the block?
Given:
M = 2.5kg. = Mass of block.
A = 17o.
Vo = 0 = Initial velocity.
V = 0.70 m/s = Final velocity.
d = 1.7 m. = Length of incline.
M*g = 2.5 * 9.8 = 24.5 N. = Wt. of block.
Fp = 24.5*sin17 = 7.16 N. = Force parallel with incline.
Fn = 24.5*cos17 = 23.43 N. = Normal force.
Fk = u*Fn = u*23.43 = 23.43u.
a. V^2 = Vo^2 + 2a*d.
0.7^2 = 0 + 2a*1.7,
a = 0.144 m/s^2.
b. Fp-Fk = M*a.
7.16-23.43u = 2.5*0.144.
u = ?
To find the acceleration of the block, we can use the equations of motion for motion along an inclined plane. In this case, we will use the equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity of the block (0.70 m/s)
- u is the initial velocity of the block (0 m/s, because it starts from rest)
- a is the acceleration of the block (to be determined)
- s is the length of the incline (1.7 m)
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the given values, we have:
a = (0.70^2 - 0^2) / (2 * 1.7)
= (0.49) / (3.4)
≈ 0.144 m/s^2
So, the acceleration of the block is approximately 0.144 m/s^2.
Now let's find the coefficient of friction between the plane and the block. The force of friction can be found using the equation:
f = μN
Where:
- f is the force of friction
- μ is the coefficient of friction (to be determined)
- N is the normal force acting on the block
The normal force can be found using the equation:
N = mgcosθ
Where:
- m is the mass of the block (2.5 kg)
- g is the acceleration due to gravity (9.8 m/s^2)
- θ is the angle of incline (17.0°)
Substituting the values, we have:
N = (2.5 kg)(9.8 m/s^2)cos(17.0°)
Now, the force of friction can be written as:
f = μ(2.5 kg)(9.8 m/s^2)cos(17.0°)
Since the block is sliding down the incline, the force of friction will act against the motion. Thus, the force of friction can be written as:
f = - μ(2.5 kg)(9.8 m/s^2)cos(17.0°)
Now, we know that the force of friction can also be calculated using the equation:
f = ma
Substituting the values, we have:
- μ(2.5 kg)(9.8 m/s^2)cos(17.0°) = (2.5 kg)(0.144 m/s^2)
Simplifying the equation by canceling out the common terms, we get:
- μ(9.8 m/s^2)cos(17.0°) = (0.144 m/s^2)
Finally, we can solve for μ:
μ = (0.144 m/s^2) / (9.8 m/s^2 * cos(17.0°))
Evaluating the expression, we find:
μ ≈ 0.025
Thus, the coefficient of friction between the plane and the block is approximately 0.025.