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harkovskaia [24]
3 years ago
10

Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of th

e impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2. What is the distance between them once they stop (express your answer to the two significant figures)
Physics
1 answer:
sattari [20]3 years ago
5 0

Answer:

48 m

Explanation:

Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of the impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2.

The eastbound train

First convert km/h to m/s

(97 × 1000)/3600

97000/3600

26.944444 m/s

As the train is decelerating, final velocity V = 0 and acceleration a will be negative. Using third equation of motion

V^2 = U^2 - 2as

O = 26.944^2 - 2 × 3.5 S

726 = 7S

S = 726/7

S1 = 103.7 m

The westbound train

Convert km/h to m/s

(127×1000)/3600

127000/3600

35.2778 m/s

Using third equation of motion

V^2 = U^2 - 2as

0 = 35.2778^2 - 2 × 4.2 × S

1244.52 = 8.4S

S = 1244.52/8.4

S2 = 148.2 m

S1 + S2 = 103.7 + 148.2 = 251.86

The distance between them once they stop will be

300 - 251.86 = 48.14 m

Therefore, the distance between them once they stop is 48 metres approximately.

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Put value of KE

59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5

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Wd'+Wd =  59.2 -29.23

                 = 29.96 J

Therefore, the work done by the drag force is given by 29.96 J

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An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ abov
Sav [38]

Answer:

a) W = 2635.56 J

b) Wf = 423.27 J

c) c)  The Sign of the work done by the frictional force (Wf) is negative (-)

d) W=0

Explanation:

Work (W) is defined as the Scalar product of the force (F) by the distance (d) that the body travels due to this force .  

The formula for calculate the work is :

W = F*d*cosα  

Where:

W : work in Joules (J)

F : force in Newtons (N)

d: displacement in meters (m)

α  :angle that form the force (F) and displacement (d)

Known data

m =  18.8 kg : mass of the block

F= 156 N,acting at an angle θ = 31.9◦°: angle  above the horizontal

μk= 0.209 : coefficient of kinetic friction between the cart and the surface

g = 9.8 m/s²: acceleration due to gravity

d = 19.9 m : displacement of the block

Forces acting on the block

We define the x-axis in the direction parallel to the movement of the cart on the floor  and the y-axis in the direction perpendicular to it.

W: Weight of the cart  : In vertical direction  downaward

N : Normal force :  In vertical direction the upaward

F : Force applied to the block

f : Friction force: In horizontal direction

Calculated of the weight  of the block

W= m*g  =  ( 18.8 kg)*(9.8 m/s²)= 184.24 N

x-y components of the force F

Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N

Fy = Fsinθ = 156 N*sin(31.9)°  = 82.44 n

Calculated of the Normal force

Newton's second law for the  block in y direction  :

∑Fy = m*ay    ay = 0

N-W+Fy= 0

N-184.24+82,44= 0

N = 184.24-82,44 

N = 101.8 N

Calculated of the kinetic friction force (fk):

fk = μk*N = (0.209)*( 101.8)

fk = 21.27 N

a) Work done by the F=156N.

W = (Fx) *d  *cosα

W = (132.44 )*(19.9)(cos0°) (N*m)

W = 2635.56 J

b) Work done by the force of friction

Wf = (fk) *d *cos(180°)

Wf = (21.27 )*(19.9) (-1) (N*m)

Wf = - 423.27 J

Wf = 423.27 J  :magnitude

c)  The Sign of the work done by the frictional force is negative (-)

d) Work done by the Normal force

W = (N) *d *cos(90°)

W = (101.8 )*(19.9) (0) (N*m)

W = 0

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3 years ago
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