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weqwewe [10]
2 years ago
7

A car whose total mass is 800kg travelling with a uniform velocity of 20m/s suddenly observes a stationary dog 50m ahead on its

path. If the total force applied on the braking system on the car by the driver is 2000N, what would happen? ​
Physics
1 answer:
maxonik [38]2 years ago
7 0

Answer:

The driver hits the stationery dog because the applied force is less than required force

Explanation:

Kinetic energy will be given by

KE=0.5mv^{2} where m is the mass of the vehicle and v is the speed/velocity of the vehicle.

Substituting 800 Kg for m and 20 m/s for v we obtain

KE=0.5*800*(20 m/s)^{2}=160,000

Frictional force by vehicle pads is given by

Fr=\frac {KE}{d} where d is the distance moved

Substituting 160000 for KE and 50 m for d we obtain

Fr=\frac {160000}{50}=3200 N

Therefore, the vehicle hits the dog since the required force is 3200N but the driver applied only 2000 N

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A sled plus passenger with total mass m = 53.1 kg is pulled a distance d = 25.3 m across a horizontal, snow-packed surface for w
Alex Ar [27]

Answer:

Explanation:

Force of friction

F = μ mg

μ is coefficient of friction , m is mass and g is acceleration due to gravity .

If f be the force applied to pull the sled , the horizontal component of force should be equal to frictional  force

The vertical component of applied force will reduce the normal force or reaction force from the ground

Reaction force R = mg - f sin28.3

frictional force = μ R where μ is coefficient of friction

frictional force = μ x (mg - f sin28.3 )

This force should be equal to horizontal component of f

μ x (mg - f sin28.3 ) = f cos 28.3

μ x mg = f μsin28.3 + f cos 28.3

f = μ x mg / (μsin28.3 + cos 28.3 )

a )

work done by pulling force  = force x displacement

f cos28.3 x d

μ x mg d cos28.3  / (μsin28.3 + cos 28.3 )

b ) Putting the given values

= .155 x 53.1 x 9.8 x 25.3 cos28.3 / ( .155 x sin28.3 + cos 28.3 )

= 1796.76 / (.073 + .88 )

= 1885.37  J

c )

Work done by frictional force

= frictional force x displacement

=  -  μ x (mg - f sin28.3 ) x d  

= -  μ x mgd + f μsin28.3  x d

= -  μ x mgd + μsin28.3  x d x μ x mg / (μsin28.3 + cos 28.3 )

d )

Putting the values in the equation above

- .155 x 53.1 x 9.8 x 25.3 +

.155 x .474 x 25.3 x .155 x 53.1 x 9.8 /( .155 x .474 + .88)

= -2040.67 + 149.92 / .95347

= -2040.67 + 157.23

= -1883.44 J .

6 0
2 years ago
a pool ball leaves a 0.60 meter high table with an initial high table with an initial horizontal velocity of 2.4m/s. predict the
timurjin [86]

Ball leaves the table of height 0.60 m

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Now here we can say that time taken by the ball to hit the ground will be given by kinematics equation in vertical direction

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here in y direction we can say

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v_y = 0

a = 9.8 m/s^2 due to gravity

now from above equation we have

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t = 0.35 s

Now in the same time the distance that ball move in horizontal direction is given as

d = v_x* t

given that

v_x = 2.4 m/s

d = 2.4 * 0.35

d = 0.84 m

So it will move a total distance of 0.84 m where it will land

5 0
2 years ago
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