Question

PLEASE DONT ROUND ANYTHING KEEP ALL DECIMALS

A 32.5 kg box initially at rest is pushed 4.10 m along a rough, horizontal floor with a constant applied horizontal force of 145 N. If the coefficient of friction between box and floor is 0.300, find the following.

(a) the work done by the applied force
J

(b) the increase in internal energy in the box-floor system due to friction
J

(c) the work done by the normal force
J

(d) the work done by the gravitational force
J

(e) the change in kinetic energy of the box
J

(f) the final speed of the box
m/s

Answer 1

a)

F = applied force on the box = 145 N

d = displacement of the box = 4.10 m

= Angle between the applied force and displacement of the box = 0 (Since they both are in same direction)

Work done by the applied force is given as

W_app = F d Cos

W_app = (145) (4.10) Cos0

W_app = 594.5 J

b)

m = mass of the box = 32.5 kg

The weight of the box in down direction is balanced by the normal force in upward direction, hence

N = normal force = weight = mg

N = mg

= Coefficient of friction between the box and floor = 0.3

frictional force acting on the box is given as

f = N

f = mg

d = displacement of the box = 4.10 m

= Angle between the frictional force and displacement of the box = 180 (Since frictional force act in opposite direction of displacement)

work done by the frictional force is given as

W_friction = f d Cos

W_friction = mg d Cos

W_friction = (0.3) (32.5 x 9.8) (4.10) Cos180

W_friction = - 391.8 J

c)

N = normal force

d = displacement of the box = 4.10 m

= Angle between the normal force and displacement of the box = 90

work done by the normal force is given as

W_n = N d Cos

W_n = N d Cos90

W_n = 0 J (Since Cos90 = 0 )

d)

F_g = force of gravity = mg

d = displacement of the box = 4.10 m

= Angle between the gravity force and displacement of the box = 90

work done by the gravity force is given as

W_g = F_g d Cos

W_g = F_g d Cos90

W_g = 0 J (Since Cos90 = 0 )

e)

As per work-change in kinetic energy theorem

Change in kinetic energy = Net work done

KE = W_app + W_friction + W_n + W_g

KE = 594.5 + ( - 391.8) + 0 + 0

KE = 202.7 J

f)

v_o = initial speed of the box = 0 m/s

v = final speed of the box = ?

m = mass of the box = 32.5 kg

Change in kinetic energy is given as

KE = (0.5) m (v² - v_o²)

202.7 = (0.5) (32.5) (v² - 0²)

v = 3.53 m/s