WORD PROBLEMS 2

Question

  A force of 10 lbs is used to move a 15-lb box across a horizontal deck, a distance of 5 ft. If the force makes an angle of 300 with the floor, 

    a) How much work is done?

    b) What is its acceleration?

    c) how long did it take the box to move

         to this distance if it started from rest?

    d) What is its final velocity?

Answer 1

I'll assume that you meant the force makes an angle of 30° instead of "300".

I'll also assume there is no friction since there is no coefficient of friction mentioned (nor any details that would allow one to be estimated).

A sketch and free-body diagram might look like this:

F_x = Fcos(θ) = 10cos(30°) = 8.66 lb (Note: This is pounds force)

To calculate Work: W = F•d

W = F_x•d = 8.66•5 = 43.3 ft lbs

To calculate box acceleration: F = ma or a = F/m however we are given the weight of the box in pounds force so we must first calculate the mass.

To calculate mass, consider again F = ma (this time in the vertical direction) where F is the weight and "a" is the acceleration due to gravity (g) of 32.2 ft/s². So m = W/g = (15 lb_f)/(32.2 ft/s²) = 0.466 lb_f•s²/ft

Back to our calculation of box acceleration: a = F/m = (8.66 lb_f)/(0.466 lbf•s²/ft) = 18.6 ft/s²

Using the kinematic equation of motion x = 1/2at² or t = √(2x/a) = √(2•5/18.6) = 0.73 sec

Using the kinematic equation of motion v = at then v = 18.6•0.73 = 13.6 ft/s