In a block and tackle, some friction in the pulleys will reduce the mechanical advantage of the machine. To include friction in a calculation of the mechanical advantage of a block and tackle, divide the weight of the object being lifted by the weight necessary to lift it.
Hope this helps
b). The power depends on the RATE at which work is done.
Power = (Work or Energy) / (time)
So to calculate it, you have to know how much work is done AND how much time that takes.
In part (a), you calculated the amount of work it takes to lift the car from the ground to Point-A. But the question doesn't tell us anywhere how much time that takes. So there's NO WAY to calculate the power needed to do it.
The more power is used, the faster the car is lifted. The less power is used, the slower the car creeps up the first hill. If the people in the car have a lot of time to sit and wait, the car can be dragged from the ground up to Point-A with a very very very small power ... you could do it with a hamster on a treadmill. That would just take a long time, but it could be done if the power is small enough.
Without knowing the time, we can't calculate the power.
...
d). Kinetic energy = (1/2) · (mass) · (speed squared)
On the way up, the car stops when it reaches point-A.
On the way down, the car leaves point-A from "rest".
WHILE it's at point-A, it has <u><em>no speed</em></u>. So it has no (<em>zero</em>) kinetic energy.
Answer:
a) F=20287.22N
b) t=2*10^-4s
Explanation:
E=1/2*m*v^2=0.5*7.8*10^-3kg*(530m/s)^2=1095.51J
The frictional force's work must be equal than the energy to stop the bullet.
So: W=F*d=F*0.054m=1095.51J, F=20287.22N
Considering the frictional force is constant, the bullet moves with constant aceleration.
a=F/m=20287.22N/7.8*10-2kg=2.6*10^6m/s^2
then d(t)=Vt-1/2*a*t^2,
5.4*10^-2m=530m/s*t-1.3*10^6m/s^2*t^2
I will calculate the time using the cuadratic formula:
with a=1.3*10^6, b=-530, c=5.4*10^-2
t=2*10^-4s
Units is the correct answer
