Answer:
v = sqrt[2*(F*h*cot(theta)-mgh)/m]
Explanation:
Work = KE + Ug
F*r=1/2mv^2+mgh
1/2mv^2=F*r-mgh
v=sqrt[2(F*r-mgh)/m]
r=h/tan(theta)=h*cot(theta)
To solve this exercise we will use the concept related to heat loss which is mathematically given as
Where,
m = mass
= Specific Heat
Change in temperature
Replacing with our values we have that
Specific heat of mercury
Replacing
Therefore the heat lost by mercury is 0.09J
If time is the x axis and distance is the y axis then yes, in the case that time is going by but distance remains the same.
The maximum static force that can be applied is equal to the normal force*the frictional force. the normal force on the box is equal to mg since the floor is flat using 9.81m/s^2 for gravity 12kg*9.81m/s^2 = 118N multiplying the normal force by the frictional force you get a 118*.42= 49.6N so overcome the force of static friction on the box a minimum of 49.6N would need to be applied.
