A 2.5 kg wooden block slides from rest down an inclined plane that makes an angle of 30o with the horizontal. If the plane has a coefficient of kinetic friction of 0.20, what is the speed of the block after slipping a distance of 2.0 m?
m = 2.5 kg, α = 30º, s =2, k = 0.2
ΔPE =KE +W(fr)
ΔPE = m•g•Δh =m•g•s•sinα
KE = m•v²/2,
W(fr) = k•N•s= k• m•g•cosα•s
m•g•s•sinα = m•v²/2 + m•g•cosα•s,
v= sqrt{2•g•s(sinα –k• cosα} =
= sqrt{2•9.8•2(0.5 -0.2•0.866)} =
=3.58 m/s
To find the speed of the block after slipping a distance of 2.0 m, we need to use the principles of physics and equations related to motion on an inclined plane.
Here are the steps to calculate the speed:
Step 1: Determine the gravitational force acting on the block.
The gravitational force can be calculated using the formula: Fg = m * g, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the values, we get Fg = 2.5 kg * 9.8 m/s² = 24.5 N.
Step 2: Calculate the component of the gravitational force parallel to the incline.
The component of the gravitational force parallel to the incline can be found using the formula: F_parallel = Fg * sin(θ), where θ is the angle of the inclined plane.
Plugging in the values, we have F_parallel = 24.5 N * sin(30°) = 12.3 N.
Step 3: Determine the force of friction.
The force of friction can be calculated using the formula: F_friction = μ_k * F_N, where μ_k is the coefficient of kinetic friction and F_N is the normal force acting on the block.
The normal force can be found using the formula: F_N = Fg * cos(θ).
Plugging in the values, we have F_N = 24.5 N * cos(30°) = 21.2 N.
Therefore, F_friction = 0.20 * 21.2 N = 4.24 N.
Step 4: Calculate the net force.
The net force is the difference between the gravitational force component parallel to the incline and the force of friction.
Net force = F_parallel - F_friction = 12.3 N - 4.24 N = 8.06 N.
Step 5: Determine the acceleration.
The acceleration of the block can be found using Newton's second law of motion: F_net = m * a, where F_net is the net force and m is the mass of the block.
Plugging in the values, we have 8.06 N = 2.5 kg * a. Solving for a, we get a = 3.22 m/s².
Step 6: Calculate the final velocity.
We can use the equation of motion for constant acceleration: v² = u² + 2 * a * s, where v is the final velocity, u is the initial velocity (which is 0 m/s since the block starts from rest), a is the acceleration, and s is the displacement.
Plugging in the values, we get v² = 0 + 2 * 3.22 m/s² * 2.0 m = 12.88 m²/s². Taking the square root, we find the final velocity v = √(12.88 m²/s²) = 3.59 m/s.
Therefore, the speed of the block after slipping a distance of 2.0 m is 3.59 m/s.