Analyzing Force & Motion on an Inclined Plane

In summary, a 5kg block is pulled up a 20 degrees incline by a 300N/5kg tension in 10 seconds. The block travels 6000m in the 10 seconds. The rope snaps and the block continues travelling up the incline for 10 more seconds. The block travels 3 seconds faster than it would if the rope hadn't snapped.
  • #1
lemon
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?
 
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  • #2
Hi lemon! :smile:

(try using the X2 tag just above the Reply box :wink:)
lemon said:
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
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
lemon said:
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
whatever's up …

lemon said:
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
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
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
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
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
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
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
Well, it was moving at half a kilometre a second …

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

1. How does the angle of an inclined plane affect the force required to move an object up the plane?

The steeper the angle of the inclined plane, the greater the force required to move an object up the plane. This is because the weight of the object is divided into two components: the force acting perpendicular to the plane (normal force) and the force acting parallel to the plane (gravity). As the angle of the plane increases, the component of gravity acting parallel to the plane increases, requiring a greater normal force to balance it and keep the object from sliding down the plane.

2. How do you calculate the force of gravity acting on an object on an inclined plane?

The force of gravity on an object on an inclined plane can be calculated using the formula Fg = mg sinθ, where Fg is the force of gravity, m is the mass of the object, and θ is the angle of the inclined plane. This formula takes into account the component of gravity acting parallel to the plane.

3. How does friction affect the motion of an object on an inclined plane?

Friction acts in the opposite direction of motion, so it can either help or hinder the motion of an object on an inclined plane. If the force of friction is greater than the force of gravity parallel to the plane, the object will not be able to move up the plane. If the force of friction is less than the force of gravity parallel to the plane, the object will accelerate down the plane.

4. Can an object on an inclined plane be in equilibrium?

Yes, an object on an inclined plane can be in equilibrium if the forces acting on the object are balanced. This means that the force of gravity parallel to the plane is equal to the force of friction acting in the opposite direction. In this case, the object will remain at rest or move at a constant velocity.

5. How can you determine the acceleration of an object on an inclined plane?

The acceleration of an object on an inclined plane can be calculated using the formula a = (Fg sinθ - Ff) / m, where a is the acceleration, Fg is the force of gravity, Ff is the force of friction, and m is the mass of the object. This formula takes into account the net force acting on the object, which includes the component of gravity parallel to the plane and the force of friction.

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