Q 1: . A car of mass 200 kg moving at 36 km/h is brought to rest after it covered a distance of 10 m. Find the retarding force acting on the car.

Answer:

Mass of the car (m) = 200 kg Initial speed (u) = 36 km/h = 10 m/s Final velocity (v) = 0

Distance covered (S) = 10 m

v2 - u2 = 2aS 0 - 100 = 2 x a x 10 - 100 = 20 aa = - 100/20 = -5 m/s2

F = ma = 200 x 5 = -1000 N

Retarding force = 1000 N

Answer:

Mass of the car (m) = 200 kg Initial speed (u) = 36 km/h = 10 m/s Final velocity (v) = 0

Distance covered (S) = 10 m

v2 - u2 = 2aS 0 - 100 = 2 x a x 10 - 100 = 20 aa = - 100/20 = -5 m/s2

F = ma = 200 x 5 = -1000 N

Retarding force = 1000 N

Q 2: What will be the change in acceleration of a sliding block, if its mass is doubled while a constant force is acting on it? ma = 2 ma

*Answer*:Force exerted on the block (F) = ma

Let force acting on the object when the mass is doubled be equal to F

Acceleration produced = a

a

i.e., acceleration is reduced to half.

Let force acting on the object when the mass is doubled be equal to F

_{1}i.e., Mass (m_{1}) = 2 mAcceleration produced = a

_{1}F_{1}= 2m x a_{1}Given F = F_{1}a

_{1 }= a/2i.e., acceleration is reduced to half.

_{1}

a = 2 a

_{1}Q 3: The figure below show a velocity time graph for a scooterist having a total mass of 150 kg. From the graph calculate - a) The acceleration in first 4 seconds b) The distance covered in the first 4 seconds. c) The force acting in the first 4 seconds.

*Answer*:a) The acceleration in the first four seconds is given by the slope of the graph AB.

AB = BF/AF

*=20-0/4 =5*m/s^{2}b) The distance covered in the first four seconds = area of Δ ABF

= 1/2 c) Force acting on the body in the first four seconds = ma = 150 x 5 = 750 N

Q 4: A certain force exerted for 1.2 seconds raises the speed of an object from 1.8 m/s to 4.2 m/s. Later the same force is applied for 2 seconds. How much does the velocity change in 2 seconds?

First calculate acceleration

a =

As the same force acts for the next two seconds the acceleration produced will be the same. The final

velocity in the first case will now become the initial velocity.

We have to calculate the final velocity at end of 2 seconds.

Acceleration (a) = 2 m/s

[First equation of motion]

v = u + at [First equation of motion]

*Answer*: Initial velocity (u) = 1.8 m/s. Final velocity (v) = 4.2 m/s Time (t) = 1.2 seconds

a =

*(v-u)/t = (4.2 - 1.8)/1.2=*2 m/s

^{2}

As the same force acts for the next two seconds the acceleration produced will be the same. The final

velocity in the first case will now become the initial velocity.

We have to calculate the final velocity at end of 2 seconds.

Acceleration (a) = 2 m/s

^{2 }Intial velocity (u) = 4.2 m/s, t = 2 s Final velocity (v) = ?

[First equation of motion]

v = u + at [First equation of motion]

= 4.2 + 2 x2 = 4.2 + 4 = 8.2 m/s

Change in velocity in two seconds = 8.2 - 4.2 = 4 m/s Post related to 9th Force and law of Motion. |

## No comments:

## Post a Comment