Question

Help Please..?
A 20-kg barrel is rolled up a 20-m ramp to the back of a truck whose floor is 5.0 m above the ground. What work is done in loading one barrel into the truck?
A:980J
B:19,600J
C :4900J
D :imposible to tell because velocity is not given.

Answer 1

You can solve this question in two ways:

WAY #1: Find the work done directly. You know the equation W=FΔd. The important thing to remember for this equation is that the force and distance must both be in the same direction. In this case, the distance is 20 m along the ramp. That means the force must also be along (parallel to) the ramp, so you need to find the component of the barrel's weight parallel to the ramp, which you should know is mgsinθ. Since the ramp is really a right triangle, you should also be able to see that you don't actually need the angle because sinθ=5/20, so:

\[W=F \Delta d=mg \sin \theta \Delta d=20(9.8) \left(5 \over 20 \right)(20)=980\]
WAY #2: Using conservation of energy. Energy is always conserved, so whatever work (a form of energy) you put into the system has to end up somewhere. That somewhere happens to be gravitational potential energy (U), because the barrel ends up higher than where it started. So:

\[W=U=mgh=20(9.8)(5)=980\]
Either way, you can see that the answer is A! Hope this helps!

Answer 2

Easier way is
W=mgh=20*9.8*5=?joule