Question

A truck drags a 200 kg log, initially at rest, across the (flat) ground for 10 meters, and then up a 30-degree incline. If the truck pulls with a constant force of 20,000 Newtons, and the coefficient of friction between the wood and ground is 0.2 for the entire duration of the journey, once log has traveled 10 meters along the incline, what is the log’s speed in m/s? Hint: use conservation of energy to solve this problem, using the log as your system, while considering the effect of outside forces on the total energy of the log. Note that friction acts along the level ground, as well as along the incline. a) What is the total work done by the truck) b) What is the total work done by friction? c)Find the speed of the log at the location of 10 meters along the incline.

Answer 1

The log first travels 10m horizontally and then 10m inclined. During horizontal it has two works, truck and friction. On inclined there is an additional work of weight (as height is changing)

a) Work done by truck = Fs = 20000 (10+10)

= 400000 J

b)

Work done by friction = -fs

but f is different here for horizontal and incline. Hence

f horizontal = nmg = 0.2 x 200 x 9.8 = 392N

f inclined = nmgcos@ = 0.2 x 200 x 9.8 x cos30 = 339.47N

work by friction = -f1 x 10 + -f1x10 = -10(392+ 339.47)

= -7314.7 J

c)

Befor applying conservation of energy, we need to find the third work as well. It is work done by weight. It will be negative as the lo is going up the plane

W = -mgh = -mgLsin30 = -200x9.8x10x0.5 = -9800J

Using

W = 1/2 mv^2

400000 - 7314.7 - 9800 = 1/2 x 200 x v^2

v = 61.88 m/s