Answer:
8,3375*10^-9 N
Explanation:
If the crate shown here is moving at a constant speed in a straight line and the force applied is 310 N, what is the magnitude o
f the force of friction?
A.310 N
B.0 N
C.620 N
A.310 N
B.0 N
C.620 N
1 answer:
Answer:
the answer is A.
Explanation:
Using the laws of newton:
∑F = ma
where ∑F is the sumatory of forces acting in the system, m the mass and a the acelertion of the system.
Then, if the block is moving with constant velocity, its aceleration is equal to 0, so:
∑F = m(0)
∑F = 0
It means that:
F - = 0
where F is the force applied and is the friction force. Replacing the value of F, we get:
310N - = 0
Finally, solving for :
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(a) 0.40 s
First of all, let's find the initial speed at which Jordan jumps from the ground.
The maximum height is h = 1.35 m. We can use the following equation:
where
v = 0 is the velocity at the maximum height
u is the initial velocity
is the acceleration of gravity
Solving for u,
The time needed to reach the maximum height can now be found by using the equation
Solving for t,
Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of
h' = 1.35 - 0.20 = 1.15 m
Using again the equation
we find
And the corresponding time is
So the time to go from h' to h is
And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:
(b) 0.08 s
This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:
And the corresponding time is
So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2: