A cyclist coasts up a 10.3° slope, traveling 17.0 m along the road to the top of the hill. If the cyclist's initial speed is 9.90 m/s, what is the final speed? Ignore friction and air resistance.
Any help would be great.
The increase in potential energy,
M g 17*sin10.3,
is equal to the decrease in kinetic energy,
(1/2)M[Vo^1 -Vf^2]
Vo = 9.90 m/s
Solve for final velocity Vf.
The M's cancel.
Vf^2 = Vo^2 - 2g*10*sin17.3
Solve for Vf
The person above me has the correct equation but has the values for the final answer mixed up. instead of doing 9.90^2-2(9.8)(10)sin(17.3), you should be doing 9.90^2-2(9.8)(17.3)sin(10.3)..hope that helps!!
NOTE: when you get your answer, make sure you take the square root of it to get your final answer. answer should be in m/s.
To find the final speed of the cyclist, we can use the conservation of mechanical energy. The initial kinetic energy (KE1) of the cyclist is equal to the final potential energy (PE2) at the top of the hill.
The initial kinetic energy (KE1) can be calculated using the formula:
KE1 = 1/2 * m * v1^2
Where:
m = mass of the cyclist (assumed to be constant)
v1 = initial speed of the cyclist
The final potential energy (PE2) at the top of the hill can be calculated using the formula:
PE2 = m * g * h
Where:
m = mass of the cyclist (assumed to be constant)
g = acceleration due to gravity (approximately 9.81 m/s^2)
h = height gained along the slope
Since the road is inclined at an angle of 10.3°, we can calculate the height gained (h) using trigonometry:
h = distance traveled * sin(slope angle)
h = 17.0 m * sin(10.3°)
Calculating h:
h ≈ 17.0 m * 0.1763
h ≈ 2.9991 m
Now, let's substitute the values into the formulas to find the initial kinetic energy (KE1) and the final potential energy (PE2):
KE1 = 1/2 * m * v1^2
KE1 = 0.5 * v1^2
PE2 = m * g * h
PE2 = m * 9.81 m/s^2 * 2.9991 m
Since KE1 = PE2 (according to the conservation of mechanical energy), we can find the final speed (v2) of the cyclist using the formula:
KE1 = PE2
0.5 * v1^2 = m * 9.81 m/s^2 * 2.9991 m
Solving for v2:
v2 = sqrt(2 * m * 9.81 m/s^2 * 2.9991 m)
Now we can substitute the given values and calculate the final speed (v2).
To find the final speed of the cyclist, we can use the principles of mechanical energy conservation. The mechanical energy of the cyclist at the bottom of the slope (initial state) is equal to the mechanical energy at the top of the slope (final state), neglecting friction and air resistance.
The mechanical energy is the sum of the kinetic energy (KE) and potential energy (PE):
ME = KE + PE
Initially, the cyclist has only kinetic energy due to its speed. At the top of the slope, the cyclist will have both kinetic and potential energy.
The kinetic energy is given by the formula:
KE = 0.5 * m * v^2
Where,
m is the mass of the cyclist (which we can assume to be constant)
v is the velocity (speed) of the cyclist
The potential energy is given by the formula:
PE = m * g * h
Where,
m is the mass of the cyclist
g is the acceleration due to gravity (9.8 m/s^2)
h is the vertical distance traveled along the slope
Since the cyclist coasts up the slope, there is no external work done, so the total mechanical energy remains constant:
ME_initial = ME_final
Initially, the cyclist has only kinetic energy, so the equation becomes:
0.5 * m * v_initial^2 = 0.5 * m * v_final^2 + m * g * h
We are given:
v_initial = 9.90 m/s (initial velocity of the cyclist)
h = 17.0 m (vertical distance traveled along the slope)
To solve for v_final, we rearrange the equation:
v_final^2 = v_initial^2 - 2 * g * h
Substituting the given values:
v_final^2 = (9.90 m/s)^2 - 2 * 9.8 m/s^2 * 17.0 m
v_final^2 = 97.92 m^2/s^2 - 333.2 m^2/s^2
v_final^2 = -235.28 m^2/s^2
Since the final speed cannot be negative in this context, we need to take the positive square root:
v_final = √(235.28) m/s
Calculating this value:
v_final ≈ 15.32 m/s
Therefore, the final speed of the cyclist is approximately 15.32 m/s.