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# Answer to Question #118675 in Mechanics | Relativity for Felix Coleman

Question #118675
1. A train of total mass 300 tonnes (3 × 105 kg) is moving up an incline of angle θ to the horizontal, where
sin θ =
1
280 . The resistance to motion is 3000 N and the train is accelerating at 0.2 ms−2
. Find
(a) the driving force of the engine,
(b) the power exerted by the driving force at the moment when the speed is 10 ms−1
.
(c) the work done by gravity when the train has moved a distance of 28 m.
2. A 54.4 kg box is being pushed a distance of 8.27m across the floor by a force F whose magnitude is 163 N.
The force F is parallel to the displacement of the box. The coefficient of kinetic friction is 0.227. Take
g = 9.81 ms−2
. Determine the work done on the box by
(a) the applied force.
(b) the friction force.
(c) the normal force.
(d) by the force of gravity.
1
2020-05-28T11:55:59-0400

Solution:

1."m=3\\sdot10^5kg;"

"sin\\theta=0.128;"

"F_r=3000N;"

"a=0.2m\/s^2;"

"v=10m\/s;"

a) "F_d=F_r+mgsin\\theta-ma;"

"F_d=3000N+300000kg\\sdot10N\/kg\\sdot0.128-300000kg\\sdot0.2m\/s^2=327000N=327kN;"

b)"P=F_dv;"

"P=327000N\\sdot10m\/s=3270000W=3.27MW;"

c)"W=Fscos\\phi;"

"W=mgsin\\theta scos\\phi;"

"W=300000kg\\sdot10N\/kg\\sdot0.128\\sdot(-1)=-384000J="

"=384kJ;"

2."m=54.4kg;"

"s=8.27m;"

"F=163N;"

"\\mu=0.227;"

"g=9.81m\/s^2";

a)"W=Fscos\\phi";

"W=163N\\sdot8.27m\\sdot1=1348J=1.348kJ;"

b)"W_f=F_fscos\\phi" ;

"F_f=\\mu N=\\mu mg;"

"F_f=0.227\\sdot54.4kg\\sdot9.81m\/s^2=121N;"

"W_f=121N\\sdot8.27m\\sdot(-1)=-1000J=-1kJ;"

c)"W_N=Nscos\\phi" ;

"N=mg;"

"W_N=mgscos\\phi;"

"W_N=54.4kg\\sdot9.81m\/s^2\\sdot8.27m\\sdot0=0;"

d) "W_g=mgscos\\phi;"

"W_g=54.4kg\\sdot9.81m\/s^2\\sdot8.27m\\sdot0=0;"

b)"P=3.27MW;"

c)"W=381kJ;"

2. a)"W=1.348kJ;"

b)"W_f=-1kJ;"

c)"W_N=0;"

d)"W_g=0;"

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