A frictionless ramp is 4 m high and 36 m long. How much force is required to push a 180 kg box up the ramp?
a) 200N
b) 300N
c) 600N
d) 900N
Wnet= mgh = 7200 J
Now that we have the machine work is still the same but force is reduced
Wnet = Fnet x Δx
sin(theta) = 4/36 =1/9
Fnet = Fpush - Fgravity = Fpush - 180 x 10 x (1/9) = Fpush-200
Wnet = (Fpush - 200) x 36 = (36 x Fpush) - 7200
so shouldnt the Force required to push the box up the ramp be 400 instead of 200?
Learn to be the most efficient. Work must be done to move it all the way up. So you have two options, one is to pull it up directly which would require a force of 1800N applied over 4 meters=7200. Or, pull it up a distance of 36 meters which decreases your force required by 9 so it's 200. The
You're getting concepts confused. The force you overcome is gravity and since you're pulling it at constant velocity there is no acceleration. The net work is 0 because it starts with 0 Ke and ends with 0 KE. So, your force is gravity. What you did would have been valid if you were accelerating it but you're not. You had the right mindset though
🙂 Remember, net work=deltaKE. So, in essence here, you do not net work. gravity does negative work and you do positive work. It sounds odd, but despite increasing PE, no net work is done. People confuse this. Work is done by two opposing forces. That equation Fd=work is for net force when used with ma. However, it is ALWAYS applicable to any force over a distance. The sum of the individual works must be added to give Net work.
In these problems, think of it as work. If there were friction, then you would be correct and you would have to apply a greater force to get balanced out but that's not the case here. In these problems always read it carefully and see what's going on. You assumed what you did because a couple of the previous problems required that. You're on your way. EK 1001 is a great way to hammer this stuff home.
