Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
From the calculations, the power expended is 43650 W.
<h3>What is the power expended?</h3>
Now we can find the acceleration from;
v = u + at
u = 0 m/s
v = 95 km/h or 26.4 m/s
t = 6.8 s
a = ?
Now
v = at
a = v/t
a = 26.4 m/s/ 6.8 s
a = 3.88 m/s^2
Force = ma = 850-kg * 3.88 m/s^2 = 3298 N
The distance covered is obtained from;
v^2 = u^2 + 2as
v^2 = 2as
s = v^2/2a
s = (26.4)^2/2 * 3.88
s = 696.96/7.76
s = 90 m
Now;
Work = Fs
Work = 3298 N * 90 m = 296820 J
Power = 296820 J/ 6.8 s
= 43650 W
Learn more about power expended:brainly.com/question/11579192
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Answer:
<h2>The angular velocity just after collision is given as</h2><h2>
</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>
Explanation:
As per given figure we know that there is no external torque about hinge point on the system of given mass
So here we will have
now we can say
so we will have
Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass
So we can use angular momentum conservation about the hinge point
Answer:
a) I = 0 N s, b) v = -3.935 m / s, c) vf = 3.935 m / s, d) y = 0.790 m
Explanation:
a) Let's start by defining the upward direction (+ y) as positive. For this part of the exercise we must use the momentum relationship
I = ∫F dt
The force of the woman on the floor is given and by action and rection the floor exerts on the woman a force of equal magnitude, but opposite direction
I = ∫ (9200 t - 11500 t2) dt
I = 9200 t² / 2 - 11500 t³ / 3
We evaluate between the lower limit t = 0 and upper limit t = 0.800 s
I = 9200 (0.8² -0) - 11500 (0.8³ -0)
I = 5888 -5888
I = 0 N s
Directed from the floor to the woman
b) For this part we use kinematics
v² = v₀² - 2g y
v = √ (0 - 2 9.8 (-0.79))
v = 3.935 m / s
The speed direction is down
c) for this we use the relationship between momentum and the amount of movement
I = ΔP
I = m vf - m v₀
vf = (I + m v₀) / m
This is the impulse of women on the floor
vf = ( 0 + 68 (3.935)) / 68
vf = 3.935 m / s
d) let's use kinematics
v₂ = v₀² - 2gy
0 = v₀² - 2gy
y = v₀² / 2g
y = 3.935²/2 9.8
y = 0.790 m
Answer:
The mass of the copper is 3.00kg
Explanation:
