M2 - work energy and power

I've finished going through the book and theres just this that i dont understand. I'm stuck on both of these questions, and looking at the answers/going through examples doesn't seem to help, so can someone please explain to me how you do them:

1: A car of mass 0.9 tonnes is driven 200m up a slope inclined at 5 degrees to the horizontal. There is a resistance force of 100N
(i) Find the work done against gravity
(ii) Find the work done against the resistance force
(iii) If the car slows down from 12 m/s to 8m/s, what is the total work done by the engine?

2: A sledge of mass 10Kg is being pulled across level ground by a rope which makes an angle of 20 degrees with the horizontal. The tension in the rope is 80N and there is a resistance force of 14N.
(i) Find the work done by a) Tension in the rope and b) the resistance force while the sledge moves a distance of 20m.
(ii) Find the speed of the sledge after it has moved 20m when
a) it starts at rest
b) it starts at 4m/s

I know there's a lot there, but I just dont see how the answers can be arrived at. Thanks in advance!

Reply 1

Chewwy

1)
i) force*dist = 900gsin5 * 200sin5
ii) force*dist = 100 * 200
iii) force = ma = 900*(-4), add that to the sum of i) and ii)

Reply 2

dholmes_7

chewwy

1)
i) force*dist = 900gsin5 * 200sin5
ii) force*dist = 100 * 200
iii) force = ma = 900*(-4), add that to the sum of i) and ii)

You did what I did. The answer to the first 1 is apparently 900g*200sin5
II) is 20 000 J
iii) 138 000 J

I dont understand how those are the answers....

Reply 3

phys981

Don´t use sin 5 twice in the first part, you should get something like 154000

Second part is correct

Third part, work out how much KE has been lost (= 36 000J)

You know the total amount of work done (add parts 1 and 2) so subtract the loss of KE and you get 138 000 as given.

Reply 4

uk6458

2

i) work done by tension:

you need the component of the tension force that acts in the direction of the sledge's movement , (horizontal):

work done = F x d

W = (80cos20) x 20
Work done = 1503.5 J

for friction:

this time friction is allready acting horizontally, so no need to resolve:

Work done = F x d

work = 14 x 20

work done = 280J

ii)

backwards force = 14N , forwards force = 80cos 20 N

resultant forwards force = 61.2N

use F=ma

61.2 = 10a
a= 6.1 ms-2

constant acceleration:

v^2 = U^2 + 2as

V^2 = 248

V = 15.7 ms-1

whaen it starts at 4 ms-1, put u=4 :

V^2 = 16 + 248
V = 16.2 ms-1

thats probably wrong, as i am not very good at these!!! someone plese correct me! :redface: