Include weight’s along-slope component as well as resistance before applying Newton’s law.
Syllabus 4.5
Notes and key methods
For uphill motion, a driving force \(\displaystyle \frac{P}{v}\) must overcome both the stated resistance and Mg sin α. Its excess over those opposing forces is the accelerating resultant Ma. Normal reaction is not an additional along-slope resistance.
At steady uphill speed, acceleration is zero, so power is (resistance+Mg sin α) times speed. Under constant power, the force changes with speed; use the instantaneous equation at the supplied speed rather than applying constant-acceleration kinematics over an unstated interval.
Check the conditions
Choose and retain a positive direction; distinguish signed quantities from magnitudes.
Use SI units and \(\displaystyle g= \frac{10m}{s^{2}}\) wherever gravity is relevant.
Apply only the model and time interval stated in the question.
Three worked examples
Example 1 · Power and acceleration on an incline
A powered vehicle of mass 400 kg moves up a fixed straight incline. The acute angle α satisfies \(\displaystyle \sin\alpha=\frac{5}{13},\quad\cos\alpha=\frac{12}{13}\). Its engine provides constant power \(\displaystyle \frac{290787}{13}\) W, parallel to motion, and resistance is a constant 57 N down the plane. Consider the instant its speed is \(\displaystyle \frac{7m}{s}\). Use \(\displaystyle g= \frac{10m}{s^{2}}\).
Find the driving force at this instant in N.
Find the uphill acceleration at this instant in \(\displaystyle \frac{m}{s^{2}}\).
Find the power needed for steady motion uphill at the stated speed, in W.
With the original constant power, find the steady uphill speed in m/s.
Show the reasoning
(a)
Convert instantaneous power to driving force at the specified nonzero speed.
The power relation gives \(\displaystyle F= \frac{P}{v}=\)\(\displaystyle \frac{41541}{13}\) N.
Answer: \(\displaystyle \frac{41541}{13}\).
(b)
Use Newton’s second law with every along-slope force.
Subtract resistance 57 and gravity component \(\displaystyle \frac{20000}{13}\) N. Dividing the remaining 1600 N by mass gives \(\displaystyle \frac{4m}{s^{2}}\).
Answer: \(4\).
(c)
Set acceleration to zero only for the explicitly changed steady-motion condition.
At steady speed \(\displaystyle F=R+Mg \sin \alpha\), giving power \(\displaystyle \frac{145187}{13}\) W.
Answer: \(\displaystyle \frac{145187}{13}\).
(d)
Solve the constant-power equation at zero acceleration with the same opposing forces.
The steady force balance gives \(\displaystyle v= \frac{P}{R+ Mg\sin \alpha }=\)\(\displaystyle \frac{41541}{2963}\) m/s.
Answer: \(\displaystyle \frac{41541}{2963}\).
Example 2 · Power and acceleration on an incline
A powered vehicle of mass 450 kg moves up a fixed straight incline. The acute angle α satisfies \(\displaystyle \sin\alpha=\frac{8}{17},\quad\cos\alpha=\frac{15}{17}\). Its engine provides constant power \(\displaystyle \frac{356000}{17}\) W, parallel to motion, and resistance is a constant 50 N down the plane. Consider the instant its speed is \(\displaystyle \frac{8m}{s}\). Use \(\displaystyle g= \frac{10m}{s^{2}}\).
Find the driving force at this instant in N.
Find the uphill acceleration at this instant in \(\displaystyle \frac{m}{s^{2}}\).
Find the power needed for steady motion uphill at the stated speed, in W.
With the original constant power, find the steady uphill speed in m/s.
Show the reasoning
(a)
Convert instantaneous power to driving force at the specified nonzero speed.
The power relation gives \(\displaystyle F= \frac{P}{v}=\)\(\displaystyle \frac{44500}{17}\) N.
Answer: \(\displaystyle \frac{44500}{17}\).
(b)
Use Newton’s second law with every along-slope force.
Subtract resistance 50 and gravity component \(\displaystyle \frac{36000}{17}\) N. Dividing the remaining 450 N by mass gives \(\displaystyle \frac{1m}{s^{2}}\).
Answer: \(1\).
(c)
Set acceleration to zero only for the explicitly changed steady-motion condition.
At steady speed \(\displaystyle F=R+Mg \sin \alpha\), giving power \(\displaystyle \frac{294800}{17}\) W.
Answer: \(\displaystyle \frac{294800}{17}\).
(d)
Solve the constant-power equation at zero acceleration with the same opposing forces.
The steady force balance gives \(\displaystyle v= \frac{P}{R+ Mg\sin \alpha }=\)\(\displaystyle \frac{7120}{737}\) m/s.
Answer: \(\displaystyle \frac{7120}{737}\).
Example 3 · Power and acceleration on an incline
A powered vehicle of mass 500 kg moves up a fixed straight incline. The acute angle α satisfies \(\displaystyle \sin\alpha=\frac{3}{5},\quad\cos\alpha=\frac{4}{5}\). Its engine provides constant power \(36459\) W, parallel to motion, and resistance is a constant 51 N down the plane. Consider the instant its speed is \(\displaystyle \frac{9m}{s}\). Use \(\displaystyle g= \frac{10m}{s^{2}}\).
Find the driving force at this instant in N.
Find the uphill acceleration at this instant in \(\displaystyle \frac{m}{s^{2}}\).
Find the power needed for steady motion uphill at the stated speed, in W.
With the original constant power, find the steady uphill speed in m/s.
Show the reasoning
(a)
Convert instantaneous power to driving force at the specified nonzero speed.
The power relation gives \(\displaystyle F= \frac{P}{v}=\)\(4051\) N.
Answer: \(4051\).
(b)
Use Newton’s second law with every along-slope force.
Subtract resistance 51 and gravity component \(3000\) N. Dividing the remaining 1000 N by mass gives \(\displaystyle \frac{2m}{s^{2}}\).
Answer: \(2\).
(c)
Set acceleration to zero only for the explicitly changed steady-motion condition.
At steady speed \(\displaystyle F=R+Mg \sin \alpha\), giving power \(27459\) W.
Answer: \(27459\).
(d)
Solve the constant-power equation at zero acceleration with the same opposing forces.
The steady force balance gives \(\displaystyle v= \frac{P}{R+ Mg\sin \alpha }=\)\(\displaystyle \frac{4051}{339}\) m/s.
Answer: \(\displaystyle \frac{4051}{339}\).
Common mistakes
Subtract both gravity’s downslope component and the stated resistance.
Do not treat the instantaneous acceleration of a constant-power model as a constant over time.