what is the net force of the box and which direction will it move (50 N left, 50N left and 100 N right)
2 answers:
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Answer:
<em>Option b is correct: 4.1 s</em>
Explanation:
<u>Vertical Launch</u>
An object launched thrown vertically upward where air resistance is negligible, reaches its maximum height in a time t, given by the equation:
Where vo is the initial speed and g is the acceleration of gravity g=9.8 .
Once the object reaches that point, it starts a free-fall motion, whose speed is (downward) given by:
The object considered in the question is thrown with vo=25 m/s. The time taken to reach the maximum height is given by [1]:
The object starts its falling motion and at some time, it has a speed of vf=15 m/s. Let's find the time by solving [2] for t:
The total time taken by the object to go up and down is
a. This option is incorrect because it's far away from the answer.
d. This option is incorrect because it's far away from the answer.
b. This option is correct because it's a good approximation to the calculated answer.
e. This option is incorrect because it's far away from the answer.
c. This option is incorrect because it's far away from the answer.
Answer:
Minimum time interval (t2)=0.90 SECONDS
Explanation:
- coefficient of friction for employees footwear = 0.5
- coefficient of friction for typical athletic shoe = 0.810
- frictional force = coefficient of friction X acceleration due to gravity X mass of body
- Acceleration due to gravity is a constant = 9.81 m/s
- Let frictional force for employee footwear = FF1
- Let frictional force for athletic footwear =FF2
FF1 = O.5 X 9.81 X mass of body
= 4.905 x mass of body
FF2 = 0.810 X 9.81 X mass of body
= 7.9461 x mass of body
The body started from rest there by making the initial velocity zero ( u = 0)
From d= ut + 1/2 a x
- d =
x a x
.....................................i
where d= distance and it is given as 3.25m
- F =ma ...................................ii
making acceleration subject of the formula from equation ii
- a =
Making t subject of formula from equation (i)
- t=
where
-
= 4.905
=7.9461
Let
- t1 = minimum time taken for frictional force for employee foot wear
- t1 =
=1.15 seconds
- t2 =
= 0.90 seconds
THANK YOU