Question

A 161 lb block travels down a 30° inclined plane with initial velocity of 10 ft/s. If the coefficient of friction is 0.2, the total work done, in ft-lbs by all forces when the block moves through a distance of 5 feet most nearly is?

Answer 1

8418 ft lbf

Step-by-step explanation:

Let g = 32 ft/s2. The gravitational force that is parallel to the incline is

`mgsin30^o = 161* 32* 0.5 = 2576 lbf`

The normal force that acting on the block is the gravity force component that is perpendicular to the incline:

`N = mgcos30^o = 32*161*√(3)/2 \approx = 4461.8 lbf`

The friction force is the product of normal force and coefficient

`F_f = \mu N = 0.2*4461.8 = 892.4 lbf`

So the net force is force of gravity subtracted by friction force

F = 2576 - 892.4 = 1683.6 lbf

So the work by this force over 5 feet distance is

W = Fs = 1683.6*5 = 8418 ft lbf