F=Pressure*area = 2*10^5(N/m^2)*pi*1.5*10^-6(m^2) = 3*pi*10^-1 N
Work reqd.(W)= gain in potential energy = m*g*h Solve it to get W in joules
F1*Cos32= F2*Cos45 F2=145*Cos32/Cos45 = 174N Any value of F2 less than 174N would result in the net force in forward direction.
Physics (work and springs)
A) The elastic potential energy stored in the spring would convert into KE of the box. This KE would be dissipated while working against the frictional force. So work done = (1/2)Kx^2 = 800*0.5^2/2 = 100 Joules B) Frictional force F= mu*mg F = 0.4*4*10 = 16 N Work done = F*d d...
Assume: P0-pressure at 4500m above sea level. P1-pressure at sea-level P2- pressure at depth of 20.6m P2=P1+dgh = P1 + 1024*9.8*20.6 = P1+2.06x10^5 Pa .......(1) P0=P1-dgh =P1 - 1.3*9.8*4500 =P1 - 0.57x10^5 Pa .......(2) [It is assumed density of air near earth's surface i...
PEi = - GMem/4Re = KEf + PEf = KEf + [- GMem/2Re] So KEf = GMem/2Re - GMem/4Re or (1/2)m v^2 = GMem/2Re v^2 = G*Me/Re
U^2 = 2*a*s s = U^2/2*a = (16.66)^2/(2*30*9.8) = 0.47m or 47 cm
Kg is unit for mass and Newton is the unit for weight.( weight is a force W=mg) Max. wt.that can be lifted = Pressure * area = 1.01*10^5*16.7*3.14*0.123^2 N =80126 N
KE gained = PE lost (1/2)mv^2 = mgh v = sqrt(2gh)= sqrt(2*9.8*0.54)m/s
(1/2)mV^2 = (1/2)kx^2 V = x*sqrt(k/m) = 0.6*sqrt(650/0.1)m/s
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