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**Question 1: A chain consisting of five links, each with mass 100 g, is lifted vertically with a constant acceleration of 2.50 m/s ^{2}, as shown in below figure. Find,**

Force acting on the object (F) is equal to the product of mas of the (m) and acceleration of the auto mobile (a).

So, F

Weight of the automobile (W) is equal to the product of mass of the automobile (m) and acceleration due to gravity on the surface of the earth (g = 9.81 m/s

W = mg

(a) The net force on each link is the same. To obtain the net force F

F

=(0.100 kg)(2.50 m/s

=(0.250 kg.m/s

=0.250 N

To obtain the weight W of the each link, substitute 0.100 kg for m and 9.81 m/s

W = mg

=(0.100 kg)( 9.81 m/s

= (0.981 kg.m/s

=0.981 N

On each link (except the top or bottom link) there is a weight W, an upward force U from the link above, and a downward force D from the link below.

So, F

Then, U= F

= (0.250 N)+(0.981 N)+D

= 1.23 N+D

For the bottom link, D = 0.

So, U = 1.23 N+D

= 1.23 N+0

=1.23 N

For the link above,

U = 1.23 N+1.23 N

= 2.46 N

For the next link above,

U = 1.23 N+2.46 N

= 3.69 N

For the next link above,

U = 1.23 N+3.69 N

= 4.92 N

From the above observation we conclude that, the force F exerted on the top link by the agent lifting the chain would be 4.92 N.

(b) To obtain the upward force U for the top link, substitute 4.92 N for D in the equation U = 1.23 N+D,

U = 1.23 N+D

= 1.23 N+4.92 N

= 6.15 N

From the above observation we conclude that, the upward force U for the top link would be 6.15 N.

(c) The net force on each link is the same. To obtain the net forces F

F

= (0.100 kg)(2.50 m/s

= (0.250 kg.m/s

= 0.250 N

From the above observation we conclude that, the forces acting between adjacent links would be 0.250 N.

(a) Let W

The upward force due to the upward acceleration a

F

If T is the tension in the upward direction on the left side block (shown in the free body diagram), then the net force will be,

m

Let W

The downward force due to the downward acceleration a

F

If T is the tension in the upward direction on the left side block (shown in the free body diagram), then the net force will be,

m

Subtracting equation (4) from equation (2),

m

Since, W

So, W

2T = (W

= (10-lb+10-lb)

So, T = 20-lb/2

= 10-lb …… (6)

From equation (6) we observed that, the reading of the scale will be 10-lb.

(b) When a single 10-lb weight (W=10-lb) is attached to a spring scale which itself is attached to a wall, the net exerted force (F

So, F

Since the weight is attached to a spring scale which itself is attached to a wall, therefore the net force will be zero (F

Putting the value of Fnet in equation (7), we get,

W - T = 0

T = W

= 10-lb …… (8)

From equation (8) we observed that, the reading of the scale will be 10-lb.

The figure below shows the car of mass M towed by another car through a rope of breaking strength T.

Apply Newton’s second law of motion along the x axis,

T cos (27°) – Mg sin (18°) = Ma

Here a

Substitute 1200 kg for M, 9.8 m/s

a

a

= {(4098.6 N) [(1 kg.m/s

= 3.41 m/s

= 0.38 m/s

Therefore, the acceleration of the car towed up by another car with a rope of breaking strength 4.6 kN is 0.38 m/s

Substitute 0.38 m/s

x = ½ (0.38 m/s

= 10.8 m

Therefore, the distance by which the car can move in 7.5 s is 10.8 m.

Concept:

(a) Substitute 1.9 kg for m

T = (1.9 kg) (9.8 m/s

= (18.62 kg. m/s

= 18.62 N

The maximum force F for which the block stays on the floor is F = 2T.

Substitute 18.62 N for T in equation F = 2T,

F = 2 (18.62 N)

= 37.24 N

Round off to two significant figures

F = 37 N

Therefore, the maximum force F for which block with mass m2 stays on floor is 37 N.

(b) The tension in the string is related to upward force as:

T = F/2

Substitute 110 N for F,

T = F/2

= 110 N/2

= 55 N

Therefore, the tension in the string corresponding to upward force of 110 N is 55 N.

(c) Substitute 55 N for T, 1.2 kg for m

a = [55 N- (1.2 kg) (9.8 m/s2)]/1.2 kg

= {(55 N) [(1 kg.m/s

= 36 m/s

Therefore, the acceleration of the block with mass m1 is 36 m/s

N = mg

= µ

Therefore, the magnitude of maximum obtainable braking force is.

(b) When the car is on the inclined road, the normal force is not equal to the weight of the car but to its component mg cosθ, that is

(a) Substitute 0.62 for µ

F = µ

= (0.62) (1500 kg) (9.81 m/s

= (9123.3 kg.m/s

= 9123. 3 N

Round off to four significant figures,

F = 9123 N

Therefore, the maximum obtainable braking force on the level road is 9123 N.

(b) Substitute 0.62 for µ

F = µ

F = µ

= (0.62) (1500 kg) (9.81 m/s

= (9020 kg. m/s

= 9020 N

Therefore, the maximum obtainable braking force on the inclined road is 9020 N.

So applying conservation of momentum to this system, the sum of the initial momentum of the man and car will be equal to the sum of the final momentum of the man and cart.

p

Substitute, m

p

m

(m

(m

(m

(m

Or, Δv

= [w/g] v

= wv

Here, weight of the man is w and weight of the railroad flat car is W.

From the above observation we conclude that, the change in velocity of the car if the man runs to the left would be wv

(a) If the man beings to climb the ladder at a speed v (with respect to ladder), in what direction and with what speed (with respect to Earth) will the balloon move?

(b) What is the state of motion after the man stops climbing?

From definition of center of mass, one can write

(M+m) v

Substitute v

0 = Mv

v

The negative sign in the equation highlights the fact that the balloon moves in the direction opposite to that of man, which is downwards in this case.

It is important to that when the balloon moves downwards with velocity vb and the man moves upward with velocity v, the velocity with which the man appears to move relative to ground is v+v

Therefore, the velocity with which the balloon moves downward is given as

v

(b) When the man stops climbing up the ladder, the balloon also stops moving downwards and eventually come to rest.

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