1) An elevator has a mass of 2000 kg, carrying three people while each of th thr
ID: 1358407 • Letter: 1
Question
1) An elevator has a mass of 2000 kg, carrying three people while each of th three people has a mass of 80 kg. The elevator lifts them 40 m. (a) If the elevator starting from rest and at the top having a velocity of 2.0 m/s. How much energy goes into lifting the elevator and the people. (b) What is the power required to lift the elevator with the people?
2) A 20.0 kg box slides down a 12.0 m long incline at an angle of 53.0 degrees with the horizontal. A force of 30 N is applied to the box to pull it up the incline. The applied force makes an agnle of 0 degrees to the incline. If the incline has a coefficient of kinetic friction of 0.100, then what is the final velocity of the box. [use energy relations]
3) A horizontal force of 12 N is applied to a 4.0 kg box that slides on a horizontal surface. The box starts from rest moves a horizontal distane of 10 meters and obtains a velocity of 5.0 m/s. The surface has friction. The friction force is? [use energy relations]
Explanation / Answer
1)
a) energy gained = (M+m)*g*h + 0.5*(M+m)*v^2
= (2000 + 3*80)*9.8*40 + 0.5*(2000+3*80)*2^2
= 882560 J
b) POwer = F*v
= (M+3*m)*g*h*v
= (2000 + 3*80)*9.8*40*2
= 1756160 W
2) Net workdone, Wnet = -30*12 + m*g*sin(53)*12 - mue_k*m*g*cos(53)*12
= -30*12 + 20*9.8*sin(53)*12 - 0.1*20*9.8*cos(53)*12
= 1376.8 J
Apply Wnet = change in kinetic enrgy
Wnet = 0.5*m*vf^2
==> vf = sqrt(2*Wnet/m)
= sqrt(2*1376.8/20)
= 11.73 m/s
3) Wnet = 12*10 - mue_k*4*9.8*
0.5*m*vf^2 = 12*10 - mue_k*4*9.8*10
0.5*4*5^2 = 12*10 - mue_k*4*9.8*10
mue_k = (12*10 - 0.5*4*5^2)/(4*9.8*10)
= 0.178
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.