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The Newton’s second law : Force (vector ) = mass (scalar) times the acceleration (vector)

We now work in 2 D. F = m a becomes 2 equations. 1 for each of the 2 directions (X and Y)

F_x (sum of all the x-components of the forces acting on the body ) = ma_x (acceleration along the horizontal)

F_y (sum of all the x-components of the forces acting on the body ) = ma_y (acceleration along the horizontal)

 

EXERCICES:

0) write the following on index cards:

REMEMBER: to find an adjacent side, use cosine and multiply by the hypothenus To find the opposite side, use sine and multiply by the hypothenus The hypothenus is the magnitude of the vector you are observing. If the vector is the weight , the hypothenus is mg To find the direction of a vector, first trace it (to identify the quadrant), then use tan( θ ) = Fy / Fx  down = negative component, left = negative component
ALONG THE X-AXIS	ALONG THE Y-AXIS
first: Write all the X-components Px = ------- (x-component of engine force if any) Nx = 0 (normal force) Wx= -------- (if any x-component of the weight) Tx = ------------ (x-component of the tension if any) fx = ------------- (friction if any) .....	first: Write all the Y-components Ny = ---------- (normal) Py = ------------ (y-component of engine) Ty = ------------ (y-component of tension) Wy = --------(y-component of the weight) fy = 0 (usually, no friction along the y-axis) ........
second: Sum all the X-components = Net force acting along the X-axis = Fnet x	Sum all the Y-components = Fnet y
third: Fnet x = m ax     if the object is accelerating along the X-axis, ax is the acceleration, m the mass or Fnet x = 0         if the object is moving at a constant speed or is not moving at all. ax = 0	Usually the object is not accelerating along the y-axis so: Fnet y = 0 because ay = 0 (if this is not the case, Fnet y = m ay ay is the acceleration along the y-axis)

1)What thrust (P)  is needed to fire a 350kg rocket straight up with an acceleration of 8.0 m/s² ?
Hint: ay = 8m/s/s , W = - mg with g = 9.8 m/s/s , and P + W = m ay along vertical

 2)What acceleration (ax) would be given to a 7.5 kg bowling ball being swung with a propelling force of 120 N (P)  ?
Hint: P = m ax solve for ax

3) A pulled tablecloth exerts a frictional force of 0.6 N (f) on a plate with mass of 0.4kg. What is the acceleration of the plate ?(hint: f = m ax)

4) 2 forces of 50N (F1) and the other of 30N (F2) , act in opposite direction on a box as shown in the diagram. 
What is the mass of the box if its acceleration is 4.0 m/s² ? (see diagram right)
Hint: F1x = 50N , F2x = - 30N  and F1x+F2x=max

 

5) A 6 kg block being pushed across a table by a force P has an acceleration of 3.0 m/s2
A) what is the net force acting on the block. (hint: use Fnet = m a)
B) If the magnitude of P is 20N, what is the magnitude of the frictional force acting on the block ? 
(hint: write Px + Fx = max  solve for Fx. Must be negative)

6) A 5kg block is acted upon by three horizontal forces as shown in the diagram (right)
A) What is the net horizontal force acting on the block ?
(hint: Find the x-components of each force and write Fnetx = F1x + F2x +F3x
 One component is negative: F1x = -10 N, F2x = 5N and F3x = 25N)
B) What is the horizontal acceleration of the block?
Hint: Fnet = m ax

 

7) what is the weight of a 40kg mass ? (use acceleration due to gravity. Weight is the pull of gravity)

 8) At a given instant in time, a 4kg rock that has been dropped from a high cliff experiences a force of air resistance of 15N. What are the magnitude and direction of the acceleration of the rock. (do not forget the force of gravity)
Hint: make a diagram. Along the vertical the air resistance F points up and the weight W points down.  Write Fy + W = m ay

9) An upward force of 18N (F) applied via a string to lift a ball with a mass of 1.5 kg.
A) what is the net force acting upon the ball ? (
Hint: Fnet = W + F   W<0
A) What is the acceleration of the ball ?
Hints: write W + F = m ay 

 10) A child is pulling a sled (with his friend on it ) and the sled accelerates.
What is the horizontal acceleration ax of the child on the sled, with combined mass 40kg, if the friction is 60N (F in red and the force is being pulled with a force of 170N at an angle of 35 degrees with the ground (P = 170N @ 35 in green) ? See diagram.
Hints: Find the x-component of each force first:
Px = __________ (use cosine)
Fx = ____________ (negative value, friction)
Then write: Px + Fx = m ax and solve for ax.

 

11) A 2.0kg weather balloon is released and begins to rise against 6.5 N of viscous drag (D). If its buoyancy is 32N (B), What is its acceleration ?
Hints : buoyancy is up B, viscous drag is down D, weight is down W
Write By + Dy + W = m ay     with Vy <0 and W <0  solve for ay

 

11) A 2.0kg weather balloon is released and begins to rise against 6.5 N of viscous drag (D). If its buoyancy is 32N (B), What is its acceleration ?

Hints : buoyancy is up B, viscous drag is down D, weight is down W

Write By + Dy + W = m ay     with Vy <0 and W <0  solve for ay

 

12) A block (5kg) is moving down an inclined plane. The inclined plane makes a 20 degrees (A) angle with the horizontal. The forces are in red and the components of forces, if any,  are in blue The coordinate system (x,y) is in orange. 
Try without hints. (see diagram)

A) Can you label the vectors ?
Find: the weight W of the block (always pointing down)
The x-component of the weight Wx
The y-component Wy
The normal force N (recoil of the plane, pushing up)
B) write sum of x-components = m ax or Wx = m ax  and find the acceleration ax along the block.
Hints:   Wx = mg sin (A) and solve for ax
C) Find the normal force.
Hints: along the vertical, there is no acceleration so sum of y-components = 0
So Wy + N = 0  with Wy = -mg cos(A) (the component is negative)
Substitute Wy in Wy + N = 0 and solve for N.
D) optional. (for Physics lovers)
There is now a friction force F = 10N along the plane. Draw the diagram again and find the new value of the acceleration. Find the coefficient of friction µk. (F = µk N)

 

13) A sky diver is falling from a aeroplane.

 A) Name the force X and the force Y.
Hints: what force pull you toward the earth ? Air above the earth will produce ...
B) State how each force changes as the sky diver speeds down
Hints: do you think the weight vary ? What about air resistance?
C) Why does the sky diver reach a steady speed ? (terminal speed)
D) Describe and explain what happen when the sky diver opens the parachute.
Hints: suddenly the air drag will become larger than the  weight

 

14) for physics lovers. Optional.
A 2.5 kg block slides down a 25 degrees inclined place with constant acceleration. 
The block starts from rest at the top. At the bottom, its velocity is 0.65 m/s. The incline is 1.6 m long. A) what is the acceleration of the block
hint: 
For question 1) you need to use Kinematics to solve for the acceleration a (a vector)
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Here are the 3 equations of motion you need to know to survive in Kinematics:
1) d = Vi t + 0.5 a t**2    (d is the displacement, a the acceleration, Vi the initial velocity (0 if it starts at rest) and t the time)
2)   a = ( V – Vi ) /t       (V is the final velocity (when the time t elapses), Vi is the initial velocity)
3) V**2 = Vi **2  + 2 a d    (V**2 meaning   V final squared)
 
V, Vi, a and d are vectors. Can be positive or negative.
a is positive if the object is accelerating when moving toward the + direction
a is negative if the object is decelerating when moving toward the  + direction
V is positive if the object is moving toward the + direction

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