# Inclined Planes – Solutions – I have used the convention in applying Newton’s second law to make all forces, displacements, velocities and accelerations down the plane negative, and up the plane positive.  The “parallel force” (F||) is just the component of gravity that is parallel to the plane.  The Perpendicular force (Fperp) is the component of weight perpendicular to the plane.

1. A 2.15 kg block of wood is on a frictionless inclined plane that makes an angle of 35.0o with the horizontal.  a) Find F||, and Fperp  b) If the block is released on the plane, what will be its acceleration down the plane?  c) What force in what direction would prevent it from accelerating down the plane?

a)

OK, basically,

F|| = (2.15)*9.8*sin(35) = 12.085N = 12.1 N

Fperp = (2.15)*9.8*cos(35) = 17.260N = 17.3 N but since there is no friction the Fperp is not terribly useful.

b)

Since there is no friction, and no other things like jet engines, strings, or little men pushing or pulling on the block, the only force along the plane is the F|| of 12.085N down the plane (-).  So our expression of Newton’s second law becomes:

F = ma,

<-12.085 N> = (2.15 kg)a, (the parallel force is negative because down the plane is negative)

so a = -5.621 m/s/s = -5.62 m/s/s (down the plane).

c)

Since there is no friction to help the block stay on the plane, you would need to cancel out the 12.085N of weight parallel to the plane, and exert exactly that force up the plane.  Having so done, you would prevent the block from accelerating up or down the plane, but it could still move at a constant velocity in any direction.  +12.1 N up the plane

2. A 5.00 kg block of wood is on a frictionless 13 m long inclined plane that makes an angle of 47o with the horizontal.  a) Find F||, and Fperp  b) If the block is released on the plane, what will be its acceleration down the plane?  c) What time will it take to slide down the plane if it is released at the top from rest?  d) What force in what direction will make the block accelerate up the plane at 4.50 m/s/s?  e) What force in what direction will make the block accelerate down the plane at 4.50 m/s/s?

a)

Again,

F|| = (5.00)*9.8*sin(47) = 35.836N = 36 N

Fperp = (5.00)*9.8*cos(47) = 33.418N = 33 N again, not terribly useful.

b)

Since there is no friction Newton’s second law is simple.  The only force acting along the plane is the parallel component of gravity down (-) the plane.  Making down the plane negative:

<-35.836 N> = (5.00 kg)a, so a = -7.167 m/s/s = -7.2 m/s/s (down the plane).

c)

s = -13, u = 0, a = -7.167 m/s/s, use s = ut + 1/2 at2. t = 1.905 s = 1.9 s

d)

So you have some unknown force up the plane (+), and the parallel component of gravity 35.836N down (-) the plane, and the acceleration is up (+) the plane, so your expression of Newton’s Second law becomes

<+F – 35.836> = (5.00 kg)(+4.50 m/s/s) making up the plane positive.  F = 58.336N = +58 N

e)

Now we have the parallel force of 35.836  down the plane (-), and a force (F) that could be up or down the plane such that the box accelerates at 4.5 m/s/s down (-) the plane.  (I know it needs a force up the plane, as the box, unfettered, will accelerate at 7.167 m/s/s down the plane) Newton’s second law gives us

<F – 35.836 N> = (5.00 kg)(-4.50 m/s/s) where the acceleration is negative because we are accelerating down the plane, and up the plane is positive.  F = 13.336 N = +13 N.  Since it is positive, it is up the plane.

3. A block of Spam is on a frictionless inclined plane that makes an angle of 35o with the horizontal.  a) If the Spam is released on the plane, what will be its acceleration down the plane?  If the Spam is given a velocity of 5.0 m/s up the plane, b) how far up the plane will the block slide before coming back down, and c) what time will it take for the block to come back down to where it was released?

a)

OK, we don’t know the mass of the Spam, but we do know that it is tasty, and also some kind of meat.  Don’t let this bother you. (Not knowing the mass)  The mass will cancel, or the problem is unsolvable, except in terms of m.  Leave m in, and solve it that same way.

F|| = m*9.8*sin(35) = 5.621m

Fperp =m*9.8*cos(35) = 8.028m again, not terribly useful as the plane is frictionless.

Since there is no friction the parallel force is the only force acting along the plane (down -) , so Newton’s second law is very simple to express:

<-5.621m> = ma, and m cancels from all terms, giving a = -5.621m/s/s = -5.6 m/s/s (down the plane) Pretty easy, huh?

b)

Solving linear kinematics, the initial velocity is up the plane (+) and the acceleration is down (-)

s = ?, u = +5.00 m/s, a = -5.621 m/s/s, v = 0 (when it reaches the top). t = ?

Use v = u + at, t = .890 s,

s = (u+v)/2*t, s = 2.224 m = +2.2 m

c)

Since there is no friction, the acceleration is the same for the block as it moves up or down the plane, and so we can simply double the amount of time it took to go up the plane (it’s like throwing a ball up in the air) so the time to return is t = 2*.890 = 1.78 s  = 1.8 s (When there is friction, the block decelerates on the way up much more quickly as it is opposed by the parallel component of gravity and friction, and accelerates more slowly down the plane.  See problems 5-8)

4. A thing is on a frictionless 3.2 m long inclined plane that makes an angle of 47o with the horizontal.  a) If the thing is released on the plane, what will be its acceleration down the plane?  b) What time will it take to slide down the plane if it is released at the top from rest?

a)

OK, again, we don’t know the mass of the thing, so let’s leave m in, and solve it that same way.

F|| = m*9.8*sin(47) = 7.167m

Fperp =m*9.8*cos(35) = 6.684m again, not terribly useful as the plane is frictionless.

Since there is no friction, the only force is the parallel force down (-) the plane. Our expression of Newton’s law is simply:

<-7.167m> = ma, and m cancels from all terms, giving a = -7.167 m/s/s = -7.2 m/s/s (down the plane)

b)

s = -3.2, u = 0, a = -7.167 m/s/s, v = ?, t = ?

Use s = ut + 1/2at2, t = .9450 s = .94 s

5. A 5.00 kg block of wood is on an inclined plane that makes an angle of 32.0o with the horizontal.  There is a static coefficient of friction of .670 and a kinetic of .340 between the block and the plane.  a) Find F||, Fperp, maximum Ffstatic, and Ffkinetic b) If the block initially at rest on the plane, will it start to slide down the plane all by itself? What is the force of static friction that exists between the plane and the block when the block is at rest on the plane?  c) What additional force is needed to make the block start to slide down the plane?  d) Once it is started down the plane, what is its acceleration down the plane?  e) What force is needed to make the block start sliding up the plane if it is initially at rest?  f) What force would make the block slide down the plane with an acceleration of 8.30 m/s/s?  g) What force would make the block slide up the plane with an acceleration of 2.55 m/s/s?

a)

F|| = (5.00)*9.8*sin(32) = 25.966 N = 26.0 N

Fperp = (5.00)*9.8*cos(32) = 41.554 N = 41.6 N– This force becomes the normal force, or the force holding the block against the plane, and is used to calculate the friction force Ff = μFN

Ffstatic(max) = μsFperp  = (.67)(41.554 N) = 27.841N = 27.8 N (This is a maximum value)

Ffkinetic = μkFperp =(.34)(41.554 N) = 14.128 N = 14.1 N

b)

If the block is sitting still on the plane, then there could be a static force of friction as high as 27.841 N holding it in place.  The F|| of 25.966 N is trying to make it slide down the plane, which is not enough, so the block sits on the plane with a force of 25.966 N down the plane, counteracted by a static frictional force of 25.966 N = +26.0 N up the plane.  (Static friction is a “up to this amount” force.  In this case, the static friction will go up to 27.841N, but no higher.) No

c)

You would need to add a force down the plane equal to the difference between the parallel component of gravity (25.966) and the maximum static frictional force.  (27.841)  So you would need 27.841-25.966 = -1.875 N = -1.9 N more (down the plane).

d)

Once it is moving down the plane, the static friction force is no longer relevant, and so we have the Ffkinetic of 14.128N up (+) the plane, (friction is in the opposite direction of motion when you are dragging something, so since we are going down the plane, it would be up the plane) and the F|| of 25.966N down (-) the plane.  Our expression of F = ma looks like

<14.128 N– 25.966 N> = (5.00 kg)a where up the plane is positive, and a = -2.368 m/s/s = -2.37 m/s/s (down the plane)

e)

You would need to overcome the static friction, and the parallel force, so you would need 27.841 + 25.966 = 53.8075 N = +55.8 N up the plane.

f)

Making up the plane positive, the acceleration is down the plane and therefore negative, and we have the kinetic friction acting up (+) the plane opposite the velocity, (assuming the block is moving – which is a good assumption since we know its acceleration) and the parallel force down (-) the plane and some applied force (F), so F = ma looks like:

<14.128 N– 25.966 N + F> = (5.00 kg)(-8.30 m/s/s), (Where up the plane is positive) and F = -29.66 N, or -29.7 N (down the plane), which makes sense because the block is wont to accelerate at only 2.368 m/s/s without any help.

g)

This is very similar to the previous problem, except now the kinetic friction force of 14.128 N is down (-) the plane, as the block is sliding up the plane, and there is the parallel force, as always, down (-) the plane.  F = ma looks like this where F is the unknown applied force:

<F – 25.966 N – 14.128 N> = (5.00 kg)(+2.55 m/s/s) (Again making up the plane positive)

so F = 52.84 N = +52.8 N up the plane.

6. A 12.0 kg block of wood is on an inclined plane that makes an angle of 55o with the horizontal.  There is a static coefficient of friction of .85 and a kinetic of .65 between the block and the plane.  a) Find F||, Fperp, maximum Ffstatic, and Ffkinetic b) If the block initially at rest on the plane, will it start to slide down the plane all by itself?  c) Once it is started down the plane, what is its acceleration down the plane?  d) What force is needed to make the block start sliding up the plane if it is initially at rest?  e) What force would make the block slide down the plane with an acceleration of 1.50 m/s/s?  f) What force would make the block slide up the plane with an acceleration of 5.4 m/s/s?

a)

F|| = (12.0)*9.8*sin(55) = 96.3323N = 96 N

Fperp = (12.0)*9.8*cos(55) = 67.4526N = 67 N– This force becomes the normal force, or the force holding the block against the plane, and is used to calculate the friction force Ff = μFN

Ffstatic(max) = μsFperp  = .85*67.4526N = 57.3347N = 57 N (This is a maximum value)

Ffkinetic = μkFperp =.65*67.4526N N = 43.8442N = 44 N

b)

Since the F|| > Ffstatic(max)  in this case it will start to slide all by itself.

c)

Once it is moving down the plane, the static friction force is no longer relevant, and so we have the Ffkinetic of 43.8442N up (+) the plane, and the F|| of 96.3323N down (-) the plane.  Our expression of F = ma looks like <43.8442N – 96.3323N> = (12.0 kg)a where up the plane is positive, and a = -4.374 m/s/s = -4.4 m/s/s (down the plane)

d)

You would need to overcome the static friction, and the parallel force, so you would need 57.3347N + 96.3323N = 153.667 N = +150 N up the plane.

e)

Making up the plane positive, as we slide down the plane, we have the kinetic friction up (+) the plane, and the parallel force (always down), so F = ma looks like:

<43.8442 N – 96.3323 N + F> = (12.0 kg)(-1.50 m/s/s), (Where up the plane is positive – the acceleration is negative as it is down the plane) and F = 34.488 N, or +34 N up the plane, which makes sense because the block will accelerate faster than 1.50 m/s/s (4.374 m/s/s) down the plane unassisted.

f)

If we are sliding up the plane, the kinetic friction now acts down (-) the plane.  Our other players along the plane are the applied force (F) and the parallel force down (-) the plane.  So F = ma looks like:

<-43.8442N – 96.3323N + F> = (12.0 kg)(+5.40 m/s/s), (Where up the plane is positive, so the acceleration is positive this time) and F = 204.977 N, or +2.0 x 102 N up the plane.

7 It takes a force of 275 N to slide a 26.0 kg box up a 50.0o inclined plane at a constant velocity. a) What is the force of friction opposing the motion?  b) What is the coefficient of friction between the box and the plane?  c) If the box were to slide freely down the plane, what would be its acceleration?  d) What force would make the box slide up the plane with an acceleration of 6.2 m/s/s?

a)

F|| = (26.0)*9.80*sin(50) = 195.1881 N

Fperp = (26.0)*9.80*cos(50) = 163.7823 N

If it is sliding up the plane with no acceleration, then the 275 N force up (+) the plane must exactly equal the parallel component of the weight down the plane (-) plus the friction force, so Newton’s second law looks like:

<+275 N – 195.1881 N + F> = (26.0 kg)(0 m/s/s) (up the plane is +, and F must be the force of friction – the acceleration is zero as it is moving at a constant velocity, and not accelerating)

F = -79.8119 N = -79.8 N (down the plane)

b)

Since this is the friction force, we can use the formula for kinetic friction to find the coefficient:

Ffkinetic = μkFperp

79.8119 N = μk(163.7823 N)

μk = .4873 = .487

c)

Were the box freely sliding, the parallel force would be acting down (-) the plane, and the friction force would oppose it and act up the plane (+) , so Newton’s second law looks like:

<-195.1881 N + 79.8119 N> = (26.0 kg)a

a = -4.4375 m/s/s = -4.44 m/s/s (down the plane)

d) Sliding up the plane, the parallel force would still be down (-) the plane, friction would be opposite the velocity and therefore down (-) the plane, and the applied force (F) would be up the plane (+).  The acceleration would be positive, as it is up the plane.:

<F - 195.1881 N - 79.8119 N> = (26.0 kg)(+6.2 m/s/s)

F = 436.2 N = +436 N up the plane

8. A piece of candy is on an inclined plane that makes an angle of 23o with the horizontal.  There is a static coefficient of friction of .72 and a kinetic of .30 between the block and the plane.  a) If the block is initially at rest on the plane, will it start to slide down the plane all by itself?  b) Once it is started down the plane, what is its acceleration down the plane?  If it is given a shove up the plane at 4.15 m/s, c) how far up the plane does it go, d) and what time does it take to get back down to where it was shoved (assuming it does not stick at the top), and e) what is its velocity when it returns?

a)

You know the drill:

F|| = m9.8*sin(23) = 3.8292m

Fperp = m9.8*cos(23) = 9.0209m– This force becomes the normal force

Ffstatic(max) = μsFperp  = .72(9.0209m) = 6.4951m (This is a maximum value)

Ffkinetic = μkFperp =.30(9.0209m) N = 2.7063m

So, the piece of candy will not start on its own as the parallel force of (3.8292m) is smaller than the maximum static force of friction (6.4951m). No

b)

Sliding down the plane, friction is up the plane (+), and parallel down (-), so F = ma looks like:

<-3.8292m + +2.7063m> = ma so a = -1.1229 m/s/s or -1.1 m/s/s (down the plane)

c)

As it slides up the plane, friction acts down the plane (-) opposite the motion, and the parallel (wie immer) is down (-) the plane:

<-3.8292m + -2.7063m> = ma, a = -6.5354 m/s/s

s = ?, u = +4.15 m/s, v = 0 (at top), a = -6.5354 m/s/s, t = ?

use v2 = u2 + 2as, s = 1.3176 m = +1.3 m up the plane

d)

s = 1.3176 m, u = +4.15 m/s, v = 0 (at top), a = -6.5354 m/s/s, t = ?

use v = u + at, t = .6350 s (for the trip up the plane)

Now we have to solve for the trip back down the plane. We have already figured out the acceleration down the plane:

s = -1.3176 m (i.e. we go back down the same distance we went up), u = 0, v = ?, a = -1.1229 m/s/s, t = ?

use s = ut + 1/2at2, t = 1.5319 s, so the round trip is .6350 s (up) + 1.5319 s (down) = 2.1669 s = 2.2 s

e)

The final velocity at the bottom:

s = -1.3176 m, u = 0, v = ?, a = -1.1229 m/s/s, t = 1.5319 s

Use v = u + at, v = -1.7202 = -1.7 m/s (down the plane) – it does not return with the same velocity as it has lost energy to friction.

9. A crate is sliding at 3.20 m/s down a 27.6o plane and is stopped by friction in a distance of 4.22 m.  a) What is the coefficient of friction between the crate and the plane?

First we have a cute linear kinematics problem to solve.  (Where else could we start??)

s = -4.22 m (down the plane), u = -3.20 m/s, v = 0 (stops), a = ???, t = don’t care

use v2 = u2 + 2as, a = +1.2133 m/s/s(up the plane)

Second we need to express the normal force, parallel force and kinetic friction in terms of m:

F|| = m9.8*sin(27.6) = 4.5403m

Fperp = m9.8*cos(27.6) = 8.6848m– This force becomes the normal force

Ffkinetic = μkFperp =μk(8.6848m)

Third – set up Newton’s second law.  As it slides down the plane, there is the parallel force down the plane (-) and the friction force up the plane (+) and the acceleration is up the plane (+) as it is stopping:

<-4.5403m + μk8.6848m> = m(+1.2133 m/s/s)  -  m divides out from both sides leaving:

<-4.5403 + μk8.6848> = +1.2133 m/s/s

μk = (4.5403 + 1.2133)/8.6848 = .662