As shown in the figure below, a box of mass m = 67.0 kg (initially at rest) is p
ID: 1497036 • Letter: A
Question
As shown in the figure below, a box of mass m = 67.0 kg (initially at rest) is pushed a distance d = 70.0 m across a rough warehouse floor by an applied force of F_A = 250 N directed at an angle of 30.0 degree below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the following. The work done by friction The work done by the normal force The work done by gravity. The work done by F_A. The velocity of the box after it has moved through distance d.Explanation / Answer
a)Normal Force N = mg + FA sin 30 degree
= 67*9.8 +250*0.5 = 781.6 N
Friction force Fr = uN
= 0.1*781.6
= 78.16N
work done by friction = Fr.d
= -78.16*70 J
= -5471.2 J
b)Work done by normal force = N*0 = 0 as the forceis normal to displacement
c)work done by gravity = 0 as the force is normal to displacement
d)Work done by FA = FA*d cos (-30) degree
=250*70*0.866
= 15155 J
e)Change in kinetic energy = work done by all the forces
0.5 m v^2 = -5471.2 + 15155
0.5*67*v^2 = 9683.8
v^2 = 9683.8/(0.5*67) =289
v = sqrt 289
v = 17 m/s
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