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

Question

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.

Answer 1

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) $W_g=mgscos\phi;$

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

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

b)$P=3.27MW;$

c)$W=381kJ;$

2. a)$W=1.348kJ;$

b)$W_f=-1kJ;$

c)$W_N=0;$

d)$W_g=0;$