A block starts with a speed of 18.0 m/s and slides for a distance of 2.2 m down a 40° ramp (ìk = 0.43). What is its final speed? (m/s)
Final KE = PE loss - (work done against friction)
(1/2)M V^2 = M g L sin40 - M g L cos40*0.43
L = 2.2 m. Cancel out the M's and solve for V^2.
What about the initial speed of 18.0 m/s when it starts?
To find the final speed of the block, we will need to use the principles of physics, specifically the equations of motion and the concept of work and energy.
First, let's break down the problem into known values:
Initial Speed (u) = 18.0 m/s
Distance traveled (s) = 2.2 m
Angle of the ramp (θ) = 40°
Coefficient of kinetic friction (μk) = 0.43
To find the final speed (v) of the block, we can divide this problem into two parts: the work done by the gravitational force and the work done by friction.
1. Work done by the gravitational force (W1):
The gravitational force can be resolved into two components: the force parallel to the ramp (mg*sinθ) and the force perpendicular to the ramp (mg*cosθ).
The work done by the gravitational force can be calculated using the formula:
W1 = m * g * s * sinθ
By substituting the given values, we get:
W1 = m * g * s * sin(40°)
2. Work done by friction (W2):
The work done by friction can be calculated using the formula:
W2 = μk * m * g * s
By substituting the given values, we get:
W2 = 0.43 * m * g * s
Now, we know that the total work done on the block is equal to its change in kinetic energy (ΔKE). This can be expressed as:
ΔKE = W1 + W2
The change in kinetic energy is given by:
ΔKE = (1/2) * m * (v^2 - u^2)
Where:
m = mass of the block (which we assume is constant and cancels out in the equation)
By equating ΔKE with W1 + W2, we can solve for the final speed (v):
(1/2) * m * (v^2 - u^2) = m * g * s * sin(θ) + μk * m * g * s
Simplifying and canceling out the masses:
(1/2) * (v^2 - u^2) = g * s * sin(θ) + μk * g * s
Finally, rearranging the equation to solve for v:
v = sqrt(u^2 + 2 * (g * s * sin(θ) + μk * g * s))
Substituting the given values, we can calculate the final speed (v).