Question

A 200 kg object is pulled up an incline plane that has friction as the picture above shows. The object is pulled at a constant velocity. Calculate the tension on the rope.

A. 897 N
B. 936 N
C. 852 N
D. 978 N

Answer 1

Step-by-step explanation:

the force to pull or push the block upwards (with the rope being parallel to the inclined plane) with 0 acceleration (moves with constant velocity) and therefore the tension on the rope is

Ft = m×g×sin(angle) + y×m×g×cos(angle)

force of gravity force of friction

down (down)

m = the mass of the object = 200kg

g = Earth's gravity acceleration = 9.8m/s²

y = the Greek letter mu, the coefficient of friction (it does not say "static" or "kinetic", I assume kinetic because the object is moving with a constant velocity, but it does not matter here) = 0.05

angle = 25 degrees

the result is N (Newton).

so, let's put the numbers into the equation and calculate :

200×9.8×sin(25) + 0.05×200×9.8×cos(25) =

≈ 828.33 N + 88.82 N ≈ 917.15 N

the closest answer option is B (936 N).

I don't know why there is this difference.