Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a co
1 answer:
Answer:
t = 1.02 s
Explanation:
The computation of the time required is shown below:
The package speed for belt is
= 3 - 1
= 2 m/s
Moreover, the decelerative force would be acted on the block i.e u.m.g
So, the decelerative produced
= 0.2 × 9.81
= 1.962 m/s^2
And, final velocity = 0
v = u - at
here
V = 0 = final velocity
u = 2 m/s
so,
0 = 2 - 1.962 × t
t = 1.02 s
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Answer:
v = 88.89 [m/s]
Explanation:
To solve this problem we must use the principle of conservation of momentum which tells us that the initial momentum of a body plus the momentum added to that body will be equal to the final momentum of the body.
We must make up the following equation:
where:
F = force applied = 4000 [N]
t = time = 0.001 [s]
m = mass = 0.045 [kg]
v = velocity [m/s]