Answer:
The required work done is 
Explanation:
Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if '
' be the coefficient of friction, then
where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.
Since the application of force by the movers does not create any acceleration to the block, we can write
So the work done (W) in moving the crate by a distance s = 10.6 m is
Answer:
Explanation:
a ) speed of passenger = circumference / time
= 2π R / Time
= 2 x 3.14 x 50 / 60
= 5.23 m /s
b )
centrifugal force = m v² /R
= (882 /9.8 ) x 5.23² / 50
= 77.47 N
Apparent weight at the highest point
real weight - centrifugal force
= 882 - 77.47
= 804.53 N
Apparent weight at the lowest point
real weight + centrifugal force
= 882 +77.47
= 959.47 N
c ) if the passenger’s apparent weight at the highest point were zero
centrifugal force = weight
mv² /R = mg
v² = gR
= 9.8 X 50
v = 22.13 m /s
d )
apparent weight
mg - mv² / R
= 882 - (882 / 9.8 )x 22.13²/50
= 882 + 882
= 1764 N
=
A reference point would be something not on the ship which could be used to calculate distance traveled.
Answer: C.) A lighthouse on a nearby Island
First, determine the mass of the object by dividing its weight on Earth by 9.8 m/s² as shown below,
m = 250 N / 9.8 m/s² = 25.51 kg
Then, multiply the obtained mass by the acceleration due to gravity (g) on Pluto.
W (in Pluto) = (25.51 kg) x (0.61 m/s²) = 15.56 N
Therefore, the object will only weigh 15.56 N.
