QUESTION 2 Show work and answer question with unit. A. You are pushing a 150-kg
ID: 1334950 • Letter: Q
Question
QUESTION 2
Show work and answer question with unit.
A. You are pushing a 150-kg wooden crate in a straight line a distance of 4.5 m across a wooden floor at constant speed. The static and kinetic coefficients of friction are 0.42 and 0.30, respectively. What is the work done by friction on the crate?
-1,984.5J (Answer, please check)
B. You are pushing a 150-kg wooden crate in a straight line a distance of 4.5 m across a wooden floor at constant speed. The static and kinetic coefficients of friction are 0.42 and 0.30, respectively. What is the work done by you on the crate?
1,984.5J (Answer, please check)
C. You are pushing a 150-kg wooden crate in a straight line a distance of 4.5 m across a wooden floor. The crate is speeding up at 0.33 m/s2. The static and kinetic coefficients of friction are 0.42 and 0.30, respectively. What is the work done by friction on the crate?
D. You are pushing a 150-kg wooden crate in a straight line a distance of 4.5 m across a wooden floor. The crate is speeding up at 0.38 m/s2. The static and kinetic coefficients of friction are 0.42 and 0.30, respectively. What is the work done by you on the crate?
Explanation / Answer
(A)
Since the crate is moving use the kinetic coefficient of friction.
Ff = muk * m * g
Ff = 150 * 9.81 * 0.30 = 441.45 N
Work done by friction force = 441.45 * 4.5 * cos(180) = -1986.525 N-m
Work done by the frictional force = -1986.525 N-m
(B)
Since the crate is moving use the kinetic coefficient of friction.
Fp = muk * m * g
[Because, the only force that opposing the motion is friction and the crate is moving with uniform speed]
Fp = 150 * 9.81 * 0.30 = 441.45 N
Work done by the person = 441.45 * 4.5 * cos(0) = 1986.525 N-m
Work done by the person = 1986.525 N-m
Both the answers given by you are correct.
(C)
Since, the frictional force is related to muk, mass and acceleration to due to gravity. Even though the crate is accelerating, frictional force value won't change.
Ff = 441.45 N
Work done by friction force = 441.45 * 4.5 * cos(180) = -1986.525 N-m
Work done by the frictional force = -1986.525 N-m
(D)
Since the crate is accelerating, the force with which you push the crate must be greater than the friction force.
Net force on the crate = mass of the crate * acceleration
Net force that led to the acceleration = F – 441.45 = 150 * 0.38
As, the kinetic friction coefficient is same I used the same friction value calculated in Part A
F = 441.45 + 57 = 498.45 N
This is the force that the person exerted.
Work done by the person = 498.45 * 4.5 * cos(0) = 2243.025 J
Work done by the person = 2243.025 J