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

a = 18.28 ft/s²

Step-by-step 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 = (15)/(32.174)\ slug`

m = 0.466 slug

now,

work done is equal to change in kinetic energy

`W = (1)/(2) m (v_f^2-v_i^2)`

`7781.5 = (1)/(2)* 0.466* v_f^2`

`v_f = 182.75\ ft/s`

now, acceleration of object

`a = (v_f-v_o)/(t)`

`a = (182.75-0)/(10)`

a = 18.28 ft/s²

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