Inclined slope

  • Thread starter lemon
  • Start date
  • #1
200
0
1. A 5kg block rests on a smooth plane inclined at an angle of 20degrees to the horizontal. The block is now pulled up the plane by a rope with a tension of 300N parallel to the plane for 10s.
a) What is the acceleration during this period?
b) What is its speed at the end of this period?
c) How far does it travel during this period?
d) The rope snaps - how much further will the block travel up the slope?
e) What is the time interval between the rope snapping and the block returning to this point?




Homework Equations


F=ma
v=u+at
s=0.5(u+v)t


3.
a) F=ma. The force is given but do I need to calculate mgsin(theta). I have a force but do not have velocity so cannot use v=u+at to find acceleration for the time period. Using F=ma I get F/m=a = 300N/5kg= 60ms^-2.
b) v=u+at 0+60ms^-2x10s =600ms^-1
c) distance=speed x time = 600ms^-1 x 10s = 6000m

Am I on the right track?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi lemon! :smile:

(try using the X2 tag just above the Reply box :wink:)
a) F=ma. The force is given but do I need to calculate mgsin(theta). I have a force but do not have velocity so cannot use v=u+at to find acceleration for the time period. Using F=ma I get F/m=a = 300N/5kg= 60ms^-2.
Yes, F = ma is correct.

(F = ma is physics, v = u + at (etc) is just geometry. This part of the question is physics, the rest is geometry.)

But you need Ftotal = ma, and you've left out gravity. :wink:
b) v=u+at 0+60ms^-2x10s =600ms^-1
Correct method. :smile:
c) distance=speed x time = 600ms^-1 x 10s = 6000m
No, distance=speed x time is only for constant speed.

You need one of the standard constant acceleration equations … which one?
 
  • #3
200
0
Hello tiny-tim:
a) so to include gravity into the calculation I calculate Fsin(theta)
300Nsin(20)=103N
F-Fsin(theta)=ma
therefore, 300N-103N=5kg x a - acceleration= 197/5 =39.4ms-[/SUP]

b) v=u+at 0+39.4ms-[/SUP] x 10s = 394ms-[/1]

c) C is asking for distance, so s=1/2(u+v)t should do it, or s=ut+1/2at[t][/SUP]
first one - 1/2(0+394)10=1970m

any better?

not getting on with the html thingys very well sorry
 
  • #4
ideasrule
Homework Helper
2,266
0
Hello tiny-tim:
a) so to include gravity into the calculation I calculate Fsin(theta)
300Nsin(20)=103N
F-Fsin(theta)=ma
therefore, 300N-103N=5kg x a - acceleration= 197/5 =39.4ms-[/SUP]
It's mgsin(theta), not Fsin(theta).
b) v=u+at 0+39.4ms-[/SUP] x 10s = 394ms-[/1]

c) C is asking for distance, so s=1/2(u+v)t should do it, or s=ut+1/2at[t][/SUP]
first one - 1/2(0+394)10=1970m

any better?

not getting on with the html thingys very well sorry
Both methods are right, but your answer to a) is wrong.
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
251
whatever's up …

not getting on with the html thingys very well sorry
Hello lemon! :smile:

Your tags should look like [noparse] and [/noparse] around whatever's up. :wink:
 
  • #6
200
0
a) so to include gravity into the calculation I calculate mgsin(theta)
5x9.81(20)=16.8N
F-F1=ma
therefore, 300N-16.8N=5kg x a - acceleration= 283.2/5 =56.6ms-2

b) v=u+at 0+56.6ms-1 x 10s = 566ms-1

c) C is asking for distance, so s=1/2(u+v)t should do it, or s=ut+1/2at[t][/sup]
first one - 1/2(0+566)10=2830m

Thanx tt
 
  • #7
200
0
d) The rope snaps - how much further will the block travel up the slope?
e) What is the time interval between the rope snapping and the block returning to this point?

d) v2=u2+2as
s=(v2)-u2/2a
s=(0ms-1-566ms-1)2/2(56.6ms-1)=2830m

e) v=u+at
t=v-u/a
= (56.6ma-1-0)/56.6ms-2=10s
and back to the original point = 2 x 10s = 20s
 
  • #8
tiny-tim
Science Advisor
Homework Helper
25,832
251
d) and e) have the correct formulas, but you must use a different acceleration now (also, either a or s must be negative :wink:)
 
  • #9
200
0
ahh! Of course. If the rope has snapped then the original tension is no longer acting upon the object. Only the gravitational force remains.

d) mgsin(theta)=ma
new acceleration = -16.8N/5kg = -3.36ms-2

s=(0ms-1-566ms-1)2/2(-3.36ms-2)=-47672.0N
47700N 3s.f.
This puppy's being yanked really hard for a 5kg block!

e) t=2(566/47700N)=0.024s 2s.f.

I think I may have gone wrong here somewhere. The force seems too big and the time too short.
 
Last edited:
  • #10
tiny-tim
Science Advisor
Homework Helper
25,832
251
erm :redface: … you've used the wrong formula for e) :rolleyes:

btw, there's no need to keep writing the units (N, s, etc). …

if you make sure that you start with everything in SI units, then the result will automatically be in SI units also (that's the whole point of a system of units). :wink:
 
  • #11
200
0
Indeed. I put in the -44.7Km distance instead of acceleration. Is this figure right for distance though? It seems awfully big.

e) t=2(566/-3.36) = -336.9s
-5min 37s
 
  • #12
tiny-tim
Science Advisor
Homework Helper
25,832
251
Well, it was moving at half a kilometre a second …

that is pretty fast. :smile:
 
  • #13
200
0
Thanks for everything today Tiny-Tim
 

Related Threads on Inclined slope

  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
20
Views
3K
  • Last Post
Replies
12
Views
18K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
3
Views
3K
Top