Unit-04

Kinetics of linear

Que-1 A 50 kg block kept on the top of 15 shopping surface in pushed down the plane with an initial velocity of 20 m/s if u=0.4 determine distance travel by block and time it will take as it come to rest .

Ans - consider FBD of block

By newton’s 2nd law

Friction force =

We know

Que-2 An election being lowered into a mine shaft starts from rest and attains a speed of 10m/s within a distance of 15m. The elevator alone has a mass of 500kg and it carries a box of 600kg mass in it. Find the total tension in cables supporting the elevator , during this acceleration motion. Also find total forces between box and floor of elevator .

Ans- consider F.B.D elevator with box

We know

By newton’s 2nd law

Consider FBD of box

Let N =normal reaction exerted between before box and floor of elevator

Que-2 At a given instant the 50N block AB moving downward with speed of 1.8 m/s . Determine its speed 2 sec later . block B has a 20N weight and coefficient of friction block B and horizontal plane is 0.2 neglect the mass of pulley’s and chord use D’ alemeberts principle

Ans- consider FBD of block A

Force acting on block A tension (2T)

Weight of block A (

Resultants force acting on block A= 2T -50----1

Since block B moving downward with acceleration

Force acting on block =-----2

Equating 1 and 2

Consider FBD of block B

---4

Since block B moving with acceleration

Force acting on block =

=2.03 ----5

Equating 4 and 5

---6

We know ,

Kinematic relation between block A and B

Differentiating w.r.t

Again differentiating w.r.t

Solving 3 and 4

Speed of block A after 2 sec is given by

Que-3 Block A of 400kg mass is being pulled up the inclined plane by using block B of 80kg mass as shown in fig. Determine the acceleration of block B and tension in rope pulling block A. Take u=0.2 assume ropes are inextensible and pulley’s are small , frictionless and massless.

Ans- kinematic relation

By virtual work principle

Total virtual work done by internet forces (tension)=0

Differentiating w.r.t

Again differentiating w.r.t

Consider FBD block A

Forces acting on block A tension T acting upward . Frictional force=uR acting downward

Resultant force acting on block A =

As block moves with acceleration

Force acting on block A = 400

Force acting on block A = 400*

Equating equation 1 and equation 2

----3

Consider F.B.D for block B

Resultant force acting on block B = 2T -7848—4

As block B moves with acceleration

Force acting on block B= 800*

Equating 4 and 5

Solving 3 and 4

Que- 4 Block A has mass of 2kg and velocity of 5m/s up the plane shown in fig. V using principle of work -energy locate the rest position of block

Ans – consider FBD for block A

Friction force=

By principle of work -energy

Work done =change in K.E

Que-5 1. Determine the distance in which can moving at 90 km/h can come to rest after pointer B switched off if u between tyres and road B 0.8.

2. Determine also max allowable speed of car if it B to stop in same distance as above on ice road where u between tyre and road is 0.08

Ans – i. u=0.8

=25m/s.

By work -energy principle

Work done = change in kinetic energy

Ii. For ice road

u=0.08

By work energy principle

Work done=change in K.E

Que-6 A pile hammer weighing 15 KN drops from height of 600mm on a pile 7.5KN. How deep a single blow of hammer drive the pile if resistance to pile is 140KN assume ground resistance is constant

Ans- velocity of hammer at time of strike

=3.431m/s

Let V be the velocity of pile and hammer immediately after impact . Applying principle of conservation of momentum of system of pile and pile hammer.

Now applying work-energy equations to system

Ques 7 The system shown in fig has a rightward velocity of 8m/s. Determine velocity after 5 second. Take for surface in contact Assume pulleys to be frictionless

Ans

Consider FBD for system

Writing impulse-momentum equation