A ball is thrown straight upward on the Moon. Is the maximum height it reaches less than, equal to, or greater than the maximum
2 answers:
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Answer:
a) x = 0.098
b) x = 2.72 m
Explanation:
(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.
When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.
Then, you have:
(1)
m: mass of the person = 62kg
k: spring constant = ?
v: velocity of the person just when he touches the fire net = ?
x: elongation of the fire net = 1.4 m
Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:
vo: initial velocity = 0 m/s
g: gravitational acceleration = 9.8 m/s^2
y: height from the person jumps = 20.0m
With this value you can find the spring constant k from the equation (1):
When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:
you solve the last expression for x:
When the person is lying on the fire net the elongation of the fire net is 0.098m
b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:
Next, you calculate x from the equation (1):
The net fire is stretched 2.72 m
Answer:
the work done to lift the bucket = 3491 Joules
Explanation:
Given:
Mass of bucket = 10kg
distance the bucket is lifted = height = 11m
Weight of rope= 0.9kg/m
g= 9.8m/s²
initial mass of water = 33kg
x = height in meters above the ground
Let W = work
Using riemann sum:
the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)
=
Work done in lifting the bucket (W) = force × distance
Force (F) = mass × acceleration due to gravity
Force = 9.8 * 10 = 98N
W done by bucket = 98×11 = 1078 Joules
Work done to lift the rope:
At Height of x meters (0≤x≤11)
Mass of rope = weight of rope × change in distance
= 0.8kg/m × (11-x)m
W done = integral of (mass×g ×distance) with upper 11 and lower limit 0
W done =
Note : upper limit is 11 not 1, problem with math editor
W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0
W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]
=7.84(60.5 -0) = 474.32 Joules
Work done in lifting the water
At Height of x meters (0≤x≤11)
Rate of water leakage = 36kg ÷ 11m = kg/m
Mass of rope = weight of rope × change in distance
= kg/m × (11-x)m = 3.27kg/m × (11-x)m
W done = integral of (mass×g ×distance) with upper 11 and lower limit 0
W done =
Note : upper limit is 11 not 1, problem with math editor
W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0
W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]
= 32.046(60.5 -0) = 1938.783 Joules
the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)
the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103
the work done to lift the bucket = 3491 Joules