Question

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.

Answer 1

Answer:

a = 18.28 ft/s²

Explanation:

given,

time of force application, t= 10 s

Work = 10 Btu

mass of the object = 15 lb

acceleration, a =  ? ft/s²

1 btu = 778.15 ft.lbf

10 btu = 7781.5 ft.lbf

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

m = 0.466 slug

now,

work done  is equal to change in kinetic energy

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

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

 $v_f = 182.75\ ft/s$

now, acceleration of object

  $a = \dfrac{v_f-v_o}{t}$

  $a = \dfrac{182.75-0}{10}$

         a = 18.28 ft/s²

constant acceleration of the object is equal to 18.28 ft/s²