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
Expert's answer
2020-05-28T11:55:59-0400

Solution:

1.m=3105kg;m=3\sdot10^5kg;

sinθ=0.128;sin\theta=0.128;

Fr=3000N;F_r=3000N;

a=0.2m/s2;a=0.2m/s^2;

v=10m/s;v=10m/s;

a) Fd=Fr+mgsinθma;F_d=F_r+mgsin\theta-ma;

Fd=3000N+300000kg10N/kg0.128300000kg0.2m/s2=327000N=327kN;F_d=3000N+300000kg\sdot10N/kg\sdot0.128-300000kg\sdot0.2m/s^2=327000N=327kN;

b)P=Fdv;P=F_dv;

P=327000N10m/s=3270000W=3.27MW;P=327000N\sdot10m/s=3270000W=3.27MW;

c)W=Fscosϕ;W=Fscos\phi;

W=mgsinθscosϕ;W=mgsin\theta scos\phi;

W=300000kg10N/kg0.128(1)=384000J=W=300000kg\sdot10N/kg\sdot0.128\sdot(-1)=-384000J=

=384kJ;=384kJ;

2.m=54.4kg;m=54.4kg;

s=8.27m;s=8.27m;

F=163N;F=163N;

μ=0.227;\mu=0.227;

g=9.81m/s2g=9.81m/s^2;

a)W=FscosϕW=Fscos\phi;

W=163N8.27m1=1348J=1.348kJ;W=163N\sdot8.27m\sdot1=1348J=1.348kJ;

b)Wf=FfscosϕW_f=F_fscos\phi ;

Ff=μN=μmg;F_f=\mu N=\mu mg;

Ff=0.22754.4kg9.81m/s2=121N;F_f=0.227\sdot54.4kg\sdot9.81m/s^2=121N;

Wf=121N8.27m(1)=1000J=1kJ;W_f=121N\sdot8.27m\sdot(-1)=-1000J=-1kJ;

c)WN=NscosϕW_N=Nscos\phi ;

N=mg;N=mg;

WN=mgscosϕ;W_N=mgscos\phi;

WN=54.4kg9.81m/s28.27m0=0;W_N=54.4kg\sdot9.81m/s^2\sdot8.27m\sdot0=0;

d) Wg=mgscosϕ;W_g=mgscos\phi;

Wg=54.4kg9.81m/s28.27m0=0;W_g=54.4kg\sdot9.81m/s^2\sdot8.27m\sdot0=0;

Answer:1.a)Fd=327kN;F_d=327kN;

b)P=3.27MW;P=3.27MW;

c)W=381kJ;W=381kJ;

2. a)W=1.348kJ;W=1.348kJ;

b)Wf=1kJ;W_f=-1kJ;

c)WN=0;W_N=0;

d)Wg=0;W_g=0;




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