the increase in kinetic energy is from (equal to) the decrease in gravitational potential energy
v^2 = 2 g h = 2 * 980 * 40 sin(.1 rd)
v^2 = 2 g h = 2 * 980 * 40 sin(.1 rd)
To begin, let's determine the potential energy of the puck at the starting point on the inclined air table. The potential energy formula is:
PE = m * g * h
Where:
- PE is the potential energy
- m is the mass of the puck
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- h is the height of the puck above a reference point
Since the puck is released at the starting point, the height (h) is zero, so the potential energy is also zero.
Next, let's calculate the potential energy of the puck at the end of the 40 cm journey. The height (h) at this point is given by the vertical displacement of the puck along the line of greatest slope.
h = 0.4 m * sin(0.1 radians)
Now, let's substitute the values into the formula to calculate the potential energy at the end point. Since we know the potential energy at the starting point is zero, the total mechanical energy at the end will equal the kinetic energy.
KE = PE = m * g * h
Now, let's equate the kinetic energy formula to the potential energy formula:
(1/2) * m * v^2 = m * g * h
Where:
- KE is the kinetic energy
- m is the mass of the puck
- v is the velocity of the puck after the 40 cm journey
- g is the acceleration due to gravity
- h is the height (calculated above)
Simplifying the equation, we get:
(1/2) * v^2 = g * h
Now, let's solve for the velocity (v):
v^2 = 2 * g * h
v = sqrt(2 * g * h)
Substituting the known values into the equation and calculating the velocity:
v = sqrt(2 * 9.8 m/s^2 * 0.4 m * sin(0.1 radians))
v ≈ 0.128 m/s
Therefore, the velocity of the puck after a 40 cm journey along the line of greatest slope is approximately 0.128 m/s.