A meter stick with a mass of 0.155 kg is pivoted about one end so it can rotate without friction about a horizontal axis. The me
ter stick is held in a horizontal position and released. As it swings through the vertical, calculate:
a. change in gravitational potential energy that has occurred;
b. the angular speed of the stick;
c. the linear speed of the end of the stick opposite the axis.
d. Compare the answer in part (c) to the speed of a particle that has fallen 1.00m, starting from rest.
a. change in gravitational potential energy that has occurred;
b. the angular speed of the stick;
c. the linear speed of the end of the stick opposite the axis.
d. Compare the answer in part (c) to the speed of a particle that has fallen 1.00m, starting from rest.
1 answer:
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Answer:
Answer: <u>Height</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>2</u><u>0</u><u>4</u><u> </u><u>m</u>
Explanation:
At the highest point, it is called the maximum height.
• From third newton's equation of motion:
• At maximum height, v is zero
• u is initial speed
• g is -9.8 m/s²
• s is the height
To solve this problem we will apply the concepts related to momentum and momentum on a body. Both are equivalent values but can be found through different expressions. The impulse is the product of the Force for time while the momentum is the product between the mass and the velocity. The result of these operations yields equivalent units.
PART A ) The Impulse can be calculcated as follows
Where,
F = Force
Change in time
Replacing,
PART B) At the same time the momentum follows the conservation of momentum where:
Initial momentum= Final momentum
And the change in momentum is equal to the Impulse, then
And
There is not initial momentum then
Answer:
a= - 6.667 m/s²
Explanation:
Given that
The initial speed of the box ,u= 20 m/s
The final speed of the box ,v= 0 m/s
The distance cover by box ,s= 30 m
Lets take the acceleration of the box = a
We know that
v²= u ² + 2 a s
Now by putting the values in the above equation we get
0²=20² + 2 a x 30
a= - 6.667 m/s²
Negative sign indicates that velocity and acceleration are in opposite direction.
Therefore the acceleration of the box will be - 6.667 m/s² .