Asked by M. Yousaf Khan

Question # 1 Marks = 12
A worker pushes 25Kg crate a distance of 5m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic frication between the crate and the floor is 0.23.
(a) What magnitude of force must the worker apply?
(b) How much work is done on the crate by this force?
(c) How much work is done on the crate by frication?
(d) How much work is done on the crate by normal force?
(e) How much work is done on the crate by the gravity?
(f) What is the total work done on the crate?
Answered by Henry
Wc = mg = 25kg * 9.8N/kg = 245N. = Weight of crate.

Fc = 245N @ 0deg.
Fp = 245sin(0) = 0 = Force parallel to floor.
Fv = 245cos(0) = 245N. = Force perpendicular to floor = Normal.

Ff = u*Fv = 0.23 * 245 = 56.35N. = Force of friction.

a. Fn = Fap - Fp - Ff = ma = 0, a = 0.
Fap - 0 - 56.35 = 0,
Fap = 56.35N. = Force applied.

b. W = Fap*d = 56.35 * 5 = 281.8J.

c. W = Ff*d = 56.35 * 5 = 281.8J.

d. W = Fv*h = 245 * 0 = 0.

e. W = Fc*h = 250 * 0 = 0.
Answered by drwls
you're way smarter than I am.
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