The second floor of a house is 6m above street level. How much work is reguired to lift a 300 kg refrigerator to the second stor
1 answer:
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1) Static friction coefficient: 0.355
The crate is initially at rest. The crate remains at rest until the horizontal pushing force is less than the maximum static frictional force.
The maximum static frictional force is given by
where
is the static coefficient of friction
m = 21 kg is the mass of the crate
g = 9.8 m/s^2 is the acceleration due to gravity
The horizontal force required to set the crate in motion is 73 N: this means that this is the value of the maximum static frictional force. So we have
Using this information into the previous equation, we can find the coefficient of static friction:
2) Kinetic friction coefficient: 0.267
Now the crate is in motion: this means that the kinetic friction is acting on the crate, and its magnitude is
(1)
where
is the coefficient of kinetic friction
There is a horizontal force of
F = 55 N
pushing the crate. Moreover, the speed of the crate is constant: this means that the acceleration is zero, a = 0.
So we can write Newton's second law as
And by substituting (1), we can find the value of the coefficient of kinetic friction:
Answer:0.015 m
Explanation:
Given
distance between two slits(d)=0.050 mm
Distance between slit and screen=2.5 m
Wavelength
and distance between First order and second order
=0.015 m