A brick is dropped from a big scaffold. What is its velocity after 4.0s?
1 answer:
The velocity of the brick is 39.2 m/s downward
Explanation:
The motion of the brick is a free fall motion, since the object is affected only by the force of gravity. Therefore, it has a uniformly accelerated motion towards the ground, with constant acceleration of .
So, we can find its velocity using the suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
For the brick in this problem (taking downward as positive direction)
u = 0 (it is dropped from rest)
Therefore, its velocity after t = 4.0 s is:
Downward, because the sign is positive.
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Answer:
1400 N
Explanation:
Change in momentum equals impulse which is a product of force and time
Change in momentum is given by m(v-u)
Equating this to impulse formula then
m(v-u)=Ft
Making F the subject of the formula then
Take upward direction as positive then downwards is negative
Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then
Answer:
The kinetic energy is: 50[J]
Explanation:
The ball is having a potential energy of 100 [J], therefore
PE = [J]
The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.
In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.
When the elevation is 5 [m], we have a potential energy of
This energy is equal to the kinetic energy, therefore
Ke= 50 [J]