20W = 20 J/s
Energy expended during climbing stairs = 50 W of energy/stair = 50J/stair
For 20 stairs, Total energy = 50x20 = 1000 J
This can light bulbs for, T= 1000J/20 J/s =50 seconds
Factors affecting friction
The intensity of friction depends on following factors: i) The area involved in friction. ii) The pressure applied on the surfaces. Force = Pressure ´ Area Frictional force will increase, if the area of contact will increase or if pressure applied on the surface increased.
Methods to reduce friction
i) Polish the contact surface. ii) Put oil or grease so that it fills in the small gaps of the flat parts. iii) Use ball bearings to reduce area of contact between rotating parts.
Lubrication
Following methods can be used to reduce friction: Oil is either thin or viscous. It depends upon SAE No. of oil. (SAE means Society of Automotive Engineers). If we use very viscous oil, it does not reach all the parts. Very thin oil will flows away easily and gets wasted. Grease is used in such cases. It is generally used around ball-bearing. Normal grease or oil is never used where there is high pressure, high temperature and high speed. Special lubricants are used in such cases. In cold season the oil becomes thick and in hot season it becomes thin. Therefore selection of lubrication also depends on the season. It is always advisable to refer operating manual of the equipment before selecting the lubricant.
The energy transferred to the spring is given by:
where
k is the spring constant
x is the elongation of the spring with respect its initial length
Let's convert the data into the SI units:
so now we can use these data inside the equation ,to find the energy transferred to the spring:
Newton's laws A force cannot act alone is the THIRD LAW!
Answer:
t₁ = 0.95 s
Explanation:
In this chaos we must use the definition of Newton's second law
F = m a = m dv / dt
dv = F dt / m
Let's replace and integrate, let's take the upward direction of the plane as positive, the force is positive
dv = ∫ (3 + 2t) dt / m
v = (3 t + 2 t²/ 2) /m
Let's evaluate between the lower limit t = 0 v = -6 ft / s (going down) to the upper limit t = t and v = 0
0 - (-6) = (3 (t- 0) + (t² -0)) / m
t² + 3t -6m = 0
Let's look for the mass
W = mg
m = W / g
m = 20/32
m = 0.625 slug
Let's solve the second degree equation
t² + 3t -3.75 = 0
t = (-3 ± √ (32 + 4 1 3.75)) / 2
t = (-3 ± 4,899) / 2
t₁ = 0.95 s
t₂ = -3.95 s
We take the positive time
