A45-kg box is being pushed a distance of 6.4 m across the floor by a force p with arrow whose magnitude is 170 n. the force p with arrow is parallel to the displacement of the box. the coefficient of kinetic friction is 0.22. determine the work done on the box by each of the four forces that act on the box. be sure to include the proper plus or minus sign for the work done by each force. (the floor is level and horizontal. forces that act on the box: pushing force p, frictional force f, normal force n, and weight mg.)

a) Work done on the box by pushing force P :  W= + 1088 J

b) Work done on the box  by the weight(W) of the box: W=0

c)  Work done on the box  by normal force N : W=0

d) Work done on the box  by the frictional force f  :W = -620.92 J

Explanation:

Work is defined as the Scalar product  of force F by the distance the body travels due to this force.  

W= F*d*cosα Formula ( 1)

Where:

W:  Work done on the body by the force (J)

d :  distance the body travels (m)

α : angle formed between force and displacement (°)

The work is positive (W+) if the force has the same direction of movement of the object.  (because α=0°, cos 0°=1)

The work is negative (W-) if the force has the opposite direction of the movement of the object.   (because α=180°, cos180°=-1)

The component of the force that performs work must be parallel to the displacement (because α=0°, cos 0°=1)

Perpendicular forces to displacement do not perform work (because α=90°,cos 90°=0)

Forces that act on the box:

Pushing force: P = 170 N          

Weight  :W=  m*g= 45 kg * 9.8 m/s²= 441 N

Normal force : N

Frictional force : f    

We apply Newton's second law to calculate N

∑F = m*a    Formula  (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

∑Fy = m*ay  ,   ay=0

N-W = 0

N=W = 441 N

We apply the following formula to calculate f:

f= μk*N  Formula (2)

Where:

Friction force : f

μk: coefficient of kinetic friction

N : Normal force (N)  

Data: μk= 0.22  ,  N= 441 N

We replace data in the formula (2)

f= 0.22*441

f= 97.02 N

Calculating the work done on the box by each of the four forces that act on the box

We apply the formula (1) to calculate the work

a) Work done on the box by pushing force P

P= 170 , d=6.4 m  , α= 0°

W=  P*d*cosα =  170N*6.4m*cos0°  = 1088 N *m

W=  1088 J

b) Work done on the box  by the weight(W) of the box

W= 441 N , d=6.4 m  , α= 90°

W= 441N *6.4m*cos90°=0

W=0

c)  Work done on the box  by normal force N

N= 441 N , d=6.4 m  , α= 90°

W= 441N *6.4m*cos90°=0

W=0

d) Work done on the box  by the frictional force f

f= 97.02 N , d=6.4 m  , α= 180°

W= 441N *6.4m*cos 180°

W = -620.92 N

found in the nucleus

has mass of one amu