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Zarrin
1 month ago
15

An object initially at rest 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 15 lb. Determine the constant horizontal acceleration in ft/s2.
Physics
1 answer:
inna [3.1K]1 month ago
3 0

Answer:

a = 18.28 ft/s²

Explanation:

the values provided are:

duration of force application, t= 10 s

Work done = 10 Btu

mass of the object = 15 lb

acceleration, a =? ft/s²

1 Btu = 778.15 ft.lbf

thus, 10 Btu = 7781.5 ft.lbf

m = \dfrac{15}{32.174}\ slug

m = 0.466 slug

So,

the work is equivalent to the change in kinetic energy

W = \dfrac{1}{2} m (v_f^2-v_i^2)

7781.5 = \dfrac{1}{2}\times 0.466\times v_f^2

 v_f = 182.75\ ft/s

The acceleration of the object is therefore

  a = \dfrac{v_f-v_o}{t}

  a = \dfrac{182.75-0}{10}

         a = 18.28 ft/s²

the constant acceleration of the object is calculated to be 18.28 ft/s²

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