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Old Sep 13th 2015, 03:38 PM   #1
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Acceleration and force

I'm having problem resolving these problems, any kind of help is welcome. You don't have to solve them, just guide me to solve them.

1. An object starts from rest and accelerates at 6 m/s2 for a time of 7 sec. What distance does the object travel?

2. An object is traveling 10 m/s and then accelerates at 4 m/s2. How much time will it take to exceed 20 m/s?

3. A ball is dropped from a 5th floor building, where each floor is 3.3 m height. How much time does it take to reach the ground? without drag.

4. A block at rest exerts a (downward) force of 20 N on an inclined plane with an angle of 30. Calculate the weight of the block.

5. Calculate the stopping distance and time for a car travelling 20 m/s with a deceleration of 6 m/s2.

6. Calculate the work done to lift a 5 kg object 3 m of height?

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Old Sep 14th 2015, 05:37 AM   #2
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There is an equation that is well worth memorising:
S=ut+(1/2)at^2
Where:
S is the distance moved
t is the elapsed time during which the object has been moving
u is the initial speed (at time t=0)
a is the acceleration
t^2 indicates elapsed time squared.

Note that this equation arises from the mathematical integration of the equation:
v=u+at
for the speed of an object under constant acceleration.

This should give you the tool required for questions 1 to 3 & 5
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Last edited by MBW; Sep 15th 2015 at 02:09 AM.
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Old Oct 1st 2015, 08:08 AM   #3
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For 1, 2, 3, 5 use the kinematic equations:

v = v0 + at
x = x0 + v0t + 1/2 at^2
a(x - x0) = 1/2 (v^2 - v0^2)


Where the subscript 0 represent the initial value of the quantity.
This is for the x direction, but you can use it for y or z direction as well, by simply substituting x with y or z.

For 4,
I guess that you are given the component of the weight perpendicular to the inclined plane.
You should be able to find the weight using trigonometry.
Keep in mind that the angle that the weight makes with the component perpendicular to the plane is the same as the angle the inclined plane makes with the horizontal.

For 6,
You must balance the force of gravity to lift the object. So the force you exert is upward with magnitude mg.
And if you have a constant force that is in the direction of the displacement, the work done by the force is the product between the magnitude of the force and the displacement.

Last edited by Roller; Oct 1st 2015 at 08:39 AM.
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