A 20-kg block, initially moving 12 m/s, slides 5 meters across a rough horizontal surface from point A to point B, producing 400 J of thermal energy. It then slides along a frictionless surface from point B up the ramp to point C.
Find the speed of the block at the top of the ramp (point C). Show your work.
To find the speed of the block at the top of the ramp (point C), we need to consider the conservation of mechanical energy.
First, let's analyze the initial situation from point A to point B. The block slides across a rough horizontal surface, so there is friction involved. The work done against friction results in the generation of thermal energy.
Given:
Mass of the block (m) = 20 kg
Initial velocity (u) = 12 m/s
Distance (d) = 5 m
Thermal energy produced (Q) = 400 J
To find the work done by friction (Wfriction), we can use the equation:
Wfriction = Q
The work done by friction is given by:
Wfriction = μ * m * g * d
Where:
μ is the coefficient of friction,
g is the acceleration due to gravity (9.8 m/s^2), and
d is the distance traveled.
Rearranging the equation, we have:
μ = Wfriction / (m * g * d)
Substituting the given values, we get:
μ = 400 J / (20 kg * 9.8 m/s^2 * 5 m)
μ = 400 J / 980 J
μ = 0.408
Now that we have the coefficient of friction, we can proceed to the next part of the problem.
The block then slides along a frictionless surface from point B up the ramp to point C. Since the ramp is frictionless, there is no work done against friction. Therefore, the total mechanical energy is conserved.
The initial mechanical energy (Ei) at point B is equal to the final mechanical energy (Ef) at point C, which consists of the kinetic energy (KE) and gravitational potential energy (GPE).
Ei = Ef
At point B:
KE1 = (1/2) * m * u^2
GPE1 = 0 (since the height is not given)
At point C:
KE2 = (1/2) * m * v^2
GPE2 = m * g * h (where h is the height at point C, which is unknown)
Since the ramp is at the same height as point B (same elevation), GPE1 = GPE2. Therefore:
KE1 = KE2
Substituting the values:
(1/2) * m * u^2 = (1/2) * m * v^2
Now we can substitute the known values:
(1/2) * 20 kg * (12 m/s)^2 = (1/2) * 20 kg * v^2
Simplifying the equation:
1440 J = 10 kg * v^2
Dividing both sides by 10 kg:
v^2 = 144 m^2/s^2
Taking the square root of both sides to find v:
v = √144 m/s
v = 12 m/s
Therefore, the speed of the block at the top of the ramp (point C) is 12 m/s.