Help with Kinematic and Energy Problems

[jinx007]

I have some problems dealing with this question please help me...!

In a downhill ski-race the total distance between start and finish is 1800 M and the total vertical drop is 550 M

The weight of the skier and equipment is 900 N and the time taken for the decent is 65 S

1/ calculate the average speed??

2/ calculate the lost in gravitational potential energy of the skier..?

3/ The average resistive force acting against the skier is 250 N, calculate the work done against this resistive force??

4/ If the skier does no work, calculate

1/ The kinetic energy as he passes the finish
2/ The speed as he passes the finish

Answer

1/ average Velocity = total distance / total time 1800/65 = 27.7 m/s

2/ Using conservation of energy P.E = K.E + P.E

mg x 550 = 1/2 x m x (27.7)^2 + mgh
so i eliminate m and (5395-383.6) = 5011.1 J here i don't know if the technique is good

3/ W = f x d

250 x 1800 = 4.5 x 10^5

4/

1/ Here i don't know how to proceed

2/ against same problem

PLEASE CHECK THE ANSWER AND HELP ME TO ATTEMPT THE LAST PART...THANK IN ADVANCE

 

[AJ Bentley]

Assuming that the 1800m is measured along the slope and does not refer to horizontal distance, 1 is correct.

In 2 you are simply asked for the change in PE - why are you calculating KE?
(Does the question actually say 'lost'? It's bad English and makes it ambiguous - it should be 'loss' I think - then it's just bad physics, the energy isn't lost, simply converted.)

3 is OK

4 The kinetic energy will be the change in PE (from 2) less the work done against friction (3).
The speed can be got from KE = 1/2mv^2. since you now know how much that is.

 

[jinx007]

Assuming that the 1800m is measured along the slope and does not refer to horizontal distance, 1 is correct.

In 2 you are simply asked for the change in PE - why are you calculating KE?
(Does the question actually say 'lost'? It's bad English and makes it ambiguous - it should be 'loss' I think - then it's just bad physics, the energy isn't lost, simply converted.)

3 is OK

4 The kinetic energy will be the change in PE (from 2) less the work done against friction (3).
The speed can be got from KE = 1/2mv^2. since you now know how much that is.

awww sorry it is loss and not lost..(in fact i know that the energy is converted) But in the question it is clearly written loss..but as we know the energy is converted let's assume it is converted...

So how should i proceed..to calculate the P.E should i just simply use p= mgh or another method..??

For the two last part the question state that the skier does no work

in fact in the paper it is like that

(b) If the skier does no work, calculate:

1/ His K.E as he passes the finish

2/ His speed as he passes the finish

And each of them carry one point don't you think that the method must be in a more simplified way..?

 

[AJ Bentley]

should i just simply use p= mgh

Yes

For the two last part the question state that the skier does no work

That just means that he adds nothing by using his sticks or pushing with his feet. In other words you can ignore him except as a dead weight.

In fact, work is done against friction but not by the skier - it's just an expression we use in such cases.
e.g. When a car uses it's brakes, 'work is done against friction'. It doesn't mean the driver does the work.

Oh, and it is simple. When you've had a bit of practice at these you'll think so too.

 

[paranormal]

1/ The kinetic energy of the skier as he passes the finish line can be calculated using the formula K.E = 1/2 x m x v^2, where m is the mass of the skier and v is the speed at the finish line.

To find the mass, we can use the weight of the skier and equipment (900 N) and the acceleration due to gravity (9.8 m/s^2). So, m = 900 N / 9.8 m/s^2 = 91.8 kg.

Now, we can plug in the values to find the kinetic energy:

K.E = 1/2 x 91.8 kg x (27.7 m/s)^2 = 36,586.59 J

2/ The speed of the skier as he passes the finish line can be calculated using the formula v = √(2gh), where h is the vertical drop (550 m).

So, v = √(2 x 9.8 m/s^2 x 550 m) = 118.5 m/s

I hope this helps with your kinematic and energy problems. If you need further assistance, feel free to ask for clarification or additional help.