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?

8418 ft lbf

Explanation:

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

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

The friction force is the product of normal force and coefficient

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

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

terd muffin is the correct answer