Question

An object with an initial horizontal velocity of 20 ft/s experiences a constant horizontal acceleration due to the action of a resultant force applied for 10 s. The work of the resultant force is 10 Btu. The mass of the object is 55 lb. Determine the constant horizontal acceleration, in ft/s2.

Answer 1

Answer:

a = 7.749 ft/s²

Explanation:

First to all, we need to convert all units, so we can work better in the calculations.

The horizontal acceleration is asked in ft/s² so the units of speed will be the same. The Work is in BTU and we need to convert it in ft.lbf in order to get the acceleration and final speed in ft/s:

W = 10 BTU * 778.15 Lbf.ft / BTU = 7781.5 lbf.ft

Now, to get the acceleration we need to get the final speed of the object first. This can be done, by using the following expression:

W = ΔKe  (1)

And Ke = 1/2mV²

So Work would be:

W = 1/2 mV₂² - 1/2mV₁²

W = 1/2m(V₂² - V₁²)    (2)

Finally, we need to convert the mass in lbf too, because Work is in lbf, so:

m = 55 lb * 1 lbf.s²/ft / 32.174 lb = 1.7095 lbf.s²/ft

Now, we can calculate the final speed by solving V₂ from (2):

7781.5 = (1/2) * (1.7095) * (V₂² - 20²)

7781.5 = 0.85475 * (V₂² - 441)

7781.5/0.85475 = (V₂² - 400)

9103.83 + 400 = V₂²

V₂ = √9503.83

V₂ = 97.49 ft/s

Now that we have the speed we can calculate the acceleration:

a = V₂ - V₁ / t

Replacing we have:

a = 97.49 - 20 / 10

a = 7.749 ft/s²

Hope this helps