This discussion is now closed.

Check out other Related discussions

- Device for 6th form and GCSEs
- GCSE grades help
- step iii
- BTEC H&S Care Unit 8: Promoting Public Health Distinction Work Explained
- Grade Growth Chronicles | From C's to A's (23-24)
- A level maths UMS marks!!!
- Weighing at Sea
- nuclear physics inverse square law
- My family say I’m stupid for not wanting to spend more than £1.1k on a laptop
- the issac physics astronomy question
- Best Laptops for uni?
- Type of ipad and keyboard for uni
- 1st year of uni- iPad dilemma
- What units do I need for Edexcel IAL or IAS Further Maths
- WJEC Unit 1 Maths As Level Discussion 15 May 2024
- I think my btec results are wrong
- Calculate my final year grade
- A level chemistry help please
- chemistry aqa a-level question help
- A2 Mechanics ladder problem help

A particle P is projected up a line of greatest slope of a rough plane which is inclined at an angle x to the horizontal, where tan x = 3/4. The coefficient of friction between P and the plane is 1/2. The particle is projected from the point O with a speed of 10ms and come to instantaneous rest at the point A.

By using work-energy principles, or otherwise

a) Find to 3 sig fig the length OA.

b) Show that P will slide back down the plane.

c) Find to 3 sf, the speed P when it returns to O

By using work-energy principles, or otherwise

a) Find to 3 sig fig the length OA.

b) Show that P will slide back down the plane.

c) Find to 3 sf, the speed P when it returns to O

a)

KE at O = 0.5m 10^2

PE at A = mg (OA sin(x))

Conservation of energy gives us:

0.5m 10^2 = mg OA sin(x)

OA = 50/(g sin(x)) = 50/(g sin(arctan(3/4))) = 8.503 m

b)

mg sin(x) - friction = mg sin(x) - 0.5 mg cos(x)

= mg (sin(x) - 0.5cos(x))

= 1.96m > 0

So the force due to gravity is greater than the friction. Therefore it must slide down.

c)

PE at A = 50m

KE at O = 0.5mv^2

External work = 1.96m

Hence work-energy gives us:

50m = 0.5mv^2 + (1.96m * 8.503)

=> v = 4.08 m/s

KE at O = 0.5m 10^2

PE at A = mg (OA sin(x))

Conservation of energy gives us:

0.5m 10^2 = mg OA sin(x)

OA = 50/(g sin(x)) = 50/(g sin(arctan(3/4))) = 8.503 m

b)

mg sin(x) - friction = mg sin(x) - 0.5 mg cos(x)

= mg (sin(x) - 0.5cos(x))

= 1.96m > 0

So the force due to gravity is greater than the friction. Therefore it must slide down.

c)

PE at A = 50m

KE at O = 0.5mv^2

External work = 1.96m

Hence work-energy gives us:

50m = 0.5mv^2 + (1.96m * 8.503)

=> v = 4.08 m/s

- Device for 6th form and GCSEs
- GCSE grades help
- step iii
- BTEC H&S Care Unit 8: Promoting Public Health Distinction Work Explained
- Grade Growth Chronicles | From C's to A's (23-24)
- A level maths UMS marks!!!
- Weighing at Sea
- nuclear physics inverse square law
- My family say I’m stupid for not wanting to spend more than £1.1k on a laptop
- the issac physics astronomy question
- Best Laptops for uni?
- Type of ipad and keyboard for uni
- 1st year of uni- iPad dilemma
- What units do I need for Edexcel IAL or IAS Further Maths
- WJEC Unit 1 Maths As Level Discussion 15 May 2024
- I think my btec results are wrong
- Calculate my final year grade
- A level chemistry help please
- chemistry aqa a-level question help
- A2 Mechanics ladder problem help

Latest

Trending

Last reply 2 weeks ago

can someone please explain what principle domain is and why the answer is a not c?Maths

0

13