Question

As shown in the figure below, a box of mass

m = 68.0 kg

(initially at rest) is pushed a distance

d = 91.0 m

across a rough warehouse floor by an applied force of

F_A = 226 N

directed at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the following. (For parts (a) through (d), give your answer to the nearest multiple of 10.)

(d)

work done by the force of friction

W_f =  

(e)

Calculate the net work on the box by finding the sum of all the works done by each individual force.

W_Net =  J

(f)

Now find the net work by first finding the net force on the box, then finding the work done by this net force.

W_Net =  J

Answer 1

The total downward force on the box, N = m * g + F_A * sin
Where m is the mass of the box, FA is the applied force, and = 30 degrees.
N = 68.0 * 9.81 + 226 * sin(30)
= 780.08 N

Frictional force, Ff = * N
Where is the coefficient of kinetic friction.
Ff = 0.1 * 780.08
= 78.01 N

Work done by frictional force, Wf = Ff * d
Where d is the distance travelled by the box.
Wf = 78.01 * 91.0
= 7098.7 J
= 7100 J

e)
Work done by the applied force, W_A = F_A * cos * d
= 226 * cos(30) * 91
= 17810.7 J

Net work done, W_Net = W_A - W_f
= 17810.7 J - 7098.7
= 10711.95 J
= 10710 J

f)
The net force on the box, F_Net = = F_A * cos - Ff
= 226 * cos(30) - 78.01
= 117.7 N

The net work done, W_Net = F_Net * d
= 117.7 * 91.0
= 10711.95 J
= 10710 J