A force of 1500 pounds drags a truck 40 feet in 20 seconds. How much work is accomplished? How much power was required?
2 answers:
Answer:
The work done and power are 60000 Pounds-feet and 3000 Pounds-feet/sec.
Explanation:
Given that,
Force F = 1500 pounds
Distance d = 40 feet
Time t = 40 sec
Work done :
The work done is the product of the force and distance.
The work done is defined as,
Where, W = work
F = force
d = distance
Put the value into the formula of work
Power:
The power is equal to the work done divided by time.
The power is defined as,
Where, P = power
W = work
t = time
Put the value into the formula of power
Hence, The work done and power are 60000 Pounds-feet and 3000 Pounds-feet/sec.
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a) Time of flight: 22.6 s
To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.
The vertical position at time t is given by
where
h = 2.5 km = 2500 m is the initial height
is the initial vertical velocity of the cargo
g = 9.8 m/s^2 is the acceleration of gravity
The cargo reaches the ground when
So substituting it into the equation and solving for t, we find the time of flight of the cargo:
b) 7.5 km
The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:
So the horizontal distance travelled is
And if we substitute the time of flight,
t = 22.6 s
We find the range of the cargo: