A crane lifts a 6.500 N weight 10 m in 180 seconds. What power is used?
1 answer:
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(a) 154.5 N
Let's divide the motion of the sprinter in two parts:
- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration for a total time
During this part of the motion, he covers a distance equal to
, until he finally reaches a velocity of
. We can use the following suvat equation:
which reduces to
(1)
since u = 0.
- In the second part, he continues with constant speed , covering a distance of
in a time
. This part of the motion is a uniform motion, so we can use the equation
(2)
We also know that the total time is 10.0 s, so
Therefore substituting into the 2nd equation
From eq.(1) we find
(3)
And substituting into (2)
Solving for t,
So from (3) we find the acceleration in the first phase:
And so the average force exerted on the sprinter is
b) 14.5 m/s
The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation
where we have
u = 0
is the acceleration
is the time of the first part
Solving the equation,