A worker pushes a box of mass m = 25 kg in a straight line along a rough floor.
ID: 1351724 • Letter: A
Question
A worker pushes a box of mass m = 25 kg in a straight line along a rough floor. The applied force F has magnitude 85 N and acts downward at an angle = 14.1° with respect to the horizontal. The box is initially at rest at the position x1 = 0 m and it has speed v2 = 0.69 m/s at position x2 = 3.50 m.
(a) Use the data to calculate the coefficient of friction.
(b) What is the net work done?
J
(c) How much work is done to overcome friction?
J
(d) What is the instantaneous power generated by the worker at x = x2?
W
Explanation / Answer
From the given data,
Normal force, N = m*g + F*sin(14.1)
= 25*9.8 + 85*sin(14.1)
= 265.7 N
a) Workdone by appled force, WF = F*d*cos(14.1)
= 85*3.5*cos(14.1)
= 288.5 J
Apply Work-energy theorem
net workdone = change in kinetic enegy
WF + W_friction = 0.5*m*(v2^2 - v1^2)
265.7 + mue_K*N*d*cos(180) = 0.5*25*(0.69^2 - 0^2)
265.7 - mue_k*265.7*3.5 = 5.95
mue_k = (265.7 - 5.95)/(265.7*3.5)
= 0.279
b) Wnet = change in kinetic enrgy
= 0.5*m*(v2^2 - v1^2)
= 0.5*25*(0.69^2 - 0^2)
= 5.95 J
c) Workdone to over come the friction = mue_k*N*d
= 0.279*265.7*3.5
= 259 J
d) P = F*v
= 85*0.69
= 58.65 Watts
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