Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what horizontal force is required to just move the blcok ( take gravity as 10m/s2
)
Answer:
108 N
Explanation:
From the question,
Applying
F' = mgμ................ Equation 1
Where F' = Frictional force = horizontal force required to just move the block, m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.
From the question,
Given: m = 18 kg, μ = 0.6, g = 10 m/s²
Substitute these values into equation 1
F' = 18×0.6×10
F' = 108 N
Answer:
Explanation:
Potential energy is the energy of a body due to is virtue of rest.
Potential energy is given as mgh
g is a constant and it is 9.81m/s²
And also the mass of the body is given as 1.3kg
Now the height of the body is
He took a book to a storey building of height 26m
He still holds the book 151 cm (1.51m) above the house.
The house is on an altitude of 1609m from the sea level.
Total Ug with out the sea level is
Ug=mgh
Ug=1.3 × 9.81 ×(26+1.51)
Ug=350.84J
Then, the potential energy due to the sea level is given as
Ug=mgh
Where g = 1/6371 m/s²
Therefore
Ug=mgh
Ug=1.3 × 1/6371 ×1609
Ug=0.328J
Total energy = 0.328+350.84
Ug=351.17J
Answer:
60N
Explanation:
in this case the minimum amount of force required must be equal to the friction Force. i.e <u>Newton</u><u>'s</u><u> </u><u>first</u><u> </u><u>law</u><u> of</u><u> </u><u>mot</u><u>ion</u><u>.</u>
therefore the maximum amount of frictional force is equal to the applied force which is 60N.
because of the net force acting on the object is zero the object is in constant motion . i.e equal and opposite force must be applied so that the object is in constant velocity therefore the total frictional force must be 60N