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:
The maximum height above the point of release is 11.653 m.
Explanation:
Given that,
Mass of block = 0.221 kg
Spring constant k = 5365 N/m
Distance x = 0.097 m
We need to calculate the height
Using stored energy in spring
...(I)
Using gravitational potential energy
....(II)
Using energy of conservation
Where, k = spring constant
m = mass of the block
x = distance
g = acceleration due to gravity
Put the value in the equation
Hence, The maximum height above the point of release is 11.653 m.
Answer: The minimum acceleration for the air plane is 2.269m/s2.
Explanation: To solve such problem the equation of motion are applicable.
The initial velocity is 0 since the airplane was initially standing. We are going to use this equation
V^2=U^2+2as
33^2=0+2a (240)
a= 2.269m/s2
Answer:
The heat transferred into the system is 183.5 J.
Explanation:
The first law of thermodynamics relates the heat transfer into or out of a system to the change of internal and the work done on the system, through the following equations.
ΔU = Q - W
where;
ΔU is the change in internal energy
Q is the heat transfer into the system
W is the work done by the system
Given;
ΔU = 155 J
W = 28.5 J
Q = ?
155 = Q - 28.5
Q = 155 + 28.5
Q = 183.5 J
Therefore, the heat transferred into the system is 183.5 J.
