Question

A man pushes horizontally on a desk that rests on a rough wooden floor. the coefficient of static friction between the desk and floor is 0.750 and the coefficient of kinetic friction is 0.600. the desk's mass is 100 kg. he pushes just hard enough to get the desk moving and continues pushing with that force for 5.00 s. how much work does he do on the desk? work = answer j

Answer 1

m = mass of desk = 100 kg

g = acceleration due to gravity = 9.8 m/s²

μ = coefficient of static friction = 0.6

F = applied force

force required to start the desk moving must be greater than or equal to static frictional force

F = static frictional force

F = μ mg

inserting the values

F = (0.750) (100 x 9.8)

F = 735 N

μ' = coefficient of kinetic friction = 0.6

once the mass starts moving , kinetic frictional force acts on it , which is given as

f' = μ' mg

a = acceleration of the mass

for the desk , force equation is given as

F - f' = ma

μ mg - μ' mg = ma

a = (μ - μ') g

a = (0.750 - 0.600) (9.8)

a = 1.47 m/s²

v₀ = initial velocity of desk = 0 m/s

t = time of travel for desk = 5.00 s

d = displacement of the desk = ?

using the equation

d = v₀ t + (0.5) a t²

d = (0) (5) + (0.5) (1.47) (5)²

d = 18.375 m

work done by man is given as

W = F d

W = (735) (18.375)

W = 13505.625 J