Mechanics ;;; please need Urgent help

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Mechanics ;;; please need Urgent help!!!!

hi everyone, i'm seeking help from anyone from this forum about my mechanics problem.

it's under WORK, ENERGY AND POWER section.

A resistive force acts on a cyclist, as she free-wheels down a straight hill at constant speed. The cyclists and her machine are modelled as a particle of mass 70kg , and the resistive force as a constant force. This constant force has a magnitude 48N and asts upwards in a direction parallel to the hill. Calculate the angle of inclination of the hill to the horizontal.

The cyclist reaches the foot of the hilla t a speed of 6ms^-1 and starts to padal, travelling along a horizontal straight road. The cyclist works at a constant rate of 624 W. by modelling the resistive fprce as a constant horizontal force of magnitude 48N, calculate the acceleration of the cyclist immediately after she starts pedalling. Show that her subsequent speed on the horizontal road connot exceed 13 ms^-1.


I have the answer from the book but it doesn't explain how to get these answer. i got the first part, i just type it to get the whole question. the answers are

4.01* for first part
0.8 ms^-2 for second part


Thanks so much to everyone.
 

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  • #2
HallsofIvy
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In the first paragraph
zawhein said:
A resistive force acts on a cyclist, as she free-wheels down a straight hill at constant speed. The cyclists and her machine are modelled as a particle of mass 70kg , and the resistive force as a constant force. This constant force has a magnitude 48N and asts upwards in a direction parallel to the hill. Calculate the angle of inclination of the hill to the horizontal.
the cyclist is moving at constant speed- that is, acceleration is 0 so net force must be 0. That 0 net force comes from two sources: gravity and the resistive force. You are told that the resistive force is 48 N parallel to the ground. The total gravitational force is mg or (70 kg)(9.8 m/s2). However, you need to find the component of that parallel to the ground. Draw a right triangle having altitude (vertical side) of "length" 70*9.8 and use trigonometry to write the base length (horizontal side) as that times a trig function. Set that equal to the 48 N resistive force and solve for the angle.

For the second paragraph
The cyclist reaches the foot of the hilla t a speed of 6ms^-1 and starts to padal, travelling along a horizontal straight road. The cyclist works at a constant rate of 624 W. by modelling the resistive fprce as a constant horizontal force of magnitude 48N, calculate the acceleration of the cyclist immediately after she starts pedalling. Show that her subsequent speed on the horizontal road connot exceed 13 ms^-1.
you know the initial speed and so the initial kinetic energy. The road is level so there is no change in potential energy. All the work done by the cyclist goes into changing the kinetic energy or fighting the resistive force.
You are told that the cyclist is doing work at 624 W (that's the power: work per second) and that the resistive force is again 48 N. Since you are asked for "acceleration of the cyclist immediately after she starts pedalling.", take some very small time, [itex]\Delta t[/itex] immediately after she starts pedalling and calculate the change in kinetic energy during that time:
The work done is 624[itex]\Delta t[/itex] Joules. The work done against the resistive force is trickier- it is 48 N times the distance moved. You might use the basic kinematic formula for that: (1/2)a ([itex]\Delta t[/itex])2)+ 6([itex]\Delta t[/itex] where a is the (unknown) acceleration.
Subtract that from the work done: 624[itex]\Delta t[/itex]-(1/2)a ([itex]\Delta t[/itex])2)+ 6([itex]\Delta t[/itex] and you have the work that goes into changing the kinetic energy. If the acceleration is a, then the change in velocity is a([itex]Delta t[/itex]), the new speed is a([itex]\Delta t[/itex])+ 6 so new kinetic energy is (1/2)((70)(a([itex]\Delta t[/itex])- 62). The total work done is the sum of the change in kinetic energy and the work done against the resistance:
[tex](1/2)(70)(a\Delta t+ 6)^2- (1/2)(70)(6)^2+ 48((1/2)a(\Delta t)^2+ 6(\Delta t))= 623 (\Delta t)[/tex]
Solve that for a. It will depend on [itex]\Delta t[/itex] but the "acceleration just after starting pedalling" should be the limit as [itex]\Delta t[/itex] goes to 0.
 
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thanks :smile:
 

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