We Know That Work Done Is Given by Product of Force and Diatnce. Hence

1. (a) (i) It is given that mass of 18 kg is at rest initially on a flat ice surface and a force of 24N is applied to it. Applying newton’s 2nd law of motion, we get

Acceleration = F/m = 24/18 = 1.33 m/sec2

We know that work done is given by product of force and diatnce. Hence,

Work done = force * distance

= 24*4 = 96 Joules

(ii) Kinetic energy attained by themass will be equal to the work done on it by the force. If v is the velocity of the mass, we can equate kinetic enegy and the work done. Hence,

(1/2)mv^2 = 96

Or, v = sqrt(96*2/m) = sqrt(96*2/18) = 3.266m/sec

(b) We know that power is the rate of doing or work or power is work done in unit time. Hence, we can write

Power = work/time = force*distance/time = force*velocity

It is given that mass is initially at rest having initially velocity as zero. After the push, the velocity attained by the mass = 3.266m/sec.

Hence, power in beginning = 24*0 = 0

Power at the end of the push = 24*3.266 = 78.384W

Average power = 78.3846/2 = 39.192W

(c) We know initial velicity, final velocity and acceleration. Hence, we can use formula v=u+ft to determine time.

Hence, 3.2659=0+1.33t or t = 3.2659/1.33 = 2.46 secs

Work done is given by product of power and time; hence, work done = Power *time or time = work/power = 96/39.192 = 2.4494 secs. It is almost same as determined above.

2. (a) Here we will use the concept of conservation of momentum. It is given that mass

M = 6600 kg is moving velocity ux = 2m/sec in x-direction and a smaller mass m is moving in –ve x-direction with the velicity double of the velocity of massM. After collission, mass M keeps on moving in x-direction but with a velocity reduced to half of the earlier velocity and mass m starts moving in x-direction with a velocity, let us say, vx. Equating momentums before and after collission, we get

Mux -2mux = Mux/2 +mvx

Or, Mux/2 = m(2ux+vx) ---- (1) It is given that the collsion is elastic, kinetic energies before and after collision will be same. Hence, (1/2Mux^2+(1/2)m*4ux^2 = (1/2)M(ux^2/4)+(1/2)mvx^2

Simplifying it, we get (3/4) Mux^2 = m(vx^2-4ux^2) --- (2)

(b) We will solve (1) and (2) to determine m and vx. As suggesed in the problem, we determine M/m from (1) and (2) and equate. We get

M/m = 2(2ux+vx)/ux = 4(vx^2-4ux^2)/3ux^2

Simplifying it, we get 3ux = 2(vx-2ux )

Or, 7ux =2vx

Or, 7*2 = 2vx as ux = 2m/sec

Hence, vx = 7 m/sec

Hence, velocity of smaller truck will be 7m/sec in x –direction.

Substituting these values in (1), we get 6600 = m(2*2+7) or m = 6600/11 = 600kg.

(c) We determine relative velocity of M and m before collission as (2+4) = 6 6m/sec in – x-direction.

Relative velocity after collission = 7-1 = 6 m/sec in x-direction.

Thus, we see that relative velocities before and after collision are same but in opposite directions.

3. (a) For rigid system to be in static state, two conditions must be satsisfied , that is, it should not move in any direction or it should not rotate also either in clockwise or anticlockwise direction. Mathematically, we can write these two conditiobs as follows.

1. Sum of the forces acting in any direction should be zero. This condition is also written as the sum of the forces in x-direction, or in y-direction or in z-direction should be zero.

2. Sum of all the moments acting any point should be zero, that is, sum clockwise moments should be equal to sum of anticlockwise moments at any point.

(b) As ladder is supported along horizontal plane and vertical wall, there will be reaction forces perpendicularto the horizonatl plane and vertical wall. As a result, there will be frictional forces acting such that these forces will try to oppose the ladder movement. Frictional force along the wall will act in vertical direction and frictional force on horizontal plane in –ve x-direction. Weight of the ladder will act downwards. It will act at the middle point of the ladder as it is uniform.

Keeping these points in mind, we can make the following free body diagram to show the various forces acting on the ladder.

(c)(i) As ladder is staic, we will hace

Sum of the forces acting in horizontal direction = 0.

Or, F2 = μF1 …. (1)

Sum of the forces acting in vertical direction =0.

Or, mg = F1+μF2 …. (2)

(ii) For staic condition of the ladder,

Sum of anticlockwise moments= sum of clockwise moments at any point.

Anti-clockwise moment at ladder center

= μF2*Lcosα+F2*Lsinα+ μF1*Lsinα

Clock wise moment at ladder center = F1*Lcosα

(iii) In equilibrium,

Sum of anticlockwise moments= sum of clockwise moments at any point.

Hence, μF2*Lcosα+F2*Lsinα+ μF1*Lsinα = F1*Lcosα

(iv) Simplifying above equation, we get

(F2+ μF1)sinα = (F1- μF2)cosα

Or, tanα = sinα/cosα = (F1- μF2)/ (F2+ μF1) F2 = μF1; hence, tanα = (F1- μ^2F1)/ (2μF1)

= (1-μ^2/2μ

(d)(i) Net force acting on floor in –ve x-direction = F2-μF1

For ladder to be stationary, we should have F2= μF1. If F2> μF1, the ladder will start moving along the floor and wall. Force (mg-F1-μF2) will act in –ve y-direction.

(ii) Under static condition, tanα = (1-μ^2/(2μ)

= (1-0.65^2)/(2*0.65)

= 0.44423

Or, α = 23.95, say 24 deg

If α becomes less than 24 deg, the ladder will start moving. Hence, this is the minimum angle which the ladder should make with the floor.