Solved Examples on Laws of Motion:-

Problem 1:-

A 26-ton Navy jet as shown in the below figure requires an air speed of 280 ft/s for lift-off. Its own engine develops a thrust of 24,000 lb. The jet is to take off from an aircraft carrier with a 300-ft flight deck. What force must be exerted by the catapult of the carrier? Assume that the catapult and the jet’s engine each exert a constant force over the 300-ft takeoff distance.

Concept:-

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

So, F = ma

Weight W of the object is equal to the mass m of the object times of the free fall acceleration g.

W = mg

m = W/g

To obtain force F in terms of weight W, substitute W/g for m in the equation F = ma,

F = ma = (W/g) (a)

Time t taken by a body to travel a distance x with average velocity v_av will be,

t = x/ v_av

The deceleration a is equal to the rate of change of velocity,

a = Δv/Δt

To find the time t for the plane to travel 300 ft, substitute 300 ft for x and 140 ft /s for v_av in the equation t = x/ v_av,

t = x/ v_av = (300 ft)/(140 ft/s) = 2.14 s

To obtain the acceleration a, substitute 280 ft/s for Δv and 2.14 s for Δt in the equation

a = Δv/Δt,

a = Δv/Δt = (280 ft/s)/(2.14 s) = 130 ft/s²

F =(W/g) (a) = (52000 lb)( 130 ft/s²)/( 32 ft/s²) = 2.1×10⁵ lb

The force exerted by the catapult F_c is equal to the difference of the net force F on the plane and the thrust develop by the own engine T.

So, F_c = F-T

F_c = F-T = (2.1×10⁵ lb) - (24,000 lb) = (2.1×10⁵ lb) - (2.4×10⁴ lb) =1.86×10⁵ lb

From the above observation we conclude that, the force exerted by the catapult F_c would be 1.86×10⁵ lb.

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Problem 2:-

A 77-kg person is parachuting and experiencing a downward acceleration of 2.5 m/s² shortly after opening the parachute. The mass of the parachute is 5.2 kg. (a) Find the upward force exerted on the parachute by the air. (b) Calculate the downward  force exerted by the person on the parachute.

Concept:-

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

So, F = ma       

From equation F = ma, the acceleration (a) of the body would be,

a = F/m   

Weight W of the object is equal to the mass m of the object times of the free fall acceleration g.

W = mg

(a) The net force F_net on the system is equal to the sum of force exerted by the person and force exerted by the parachute.

So, F_net = (m_pe+m_pa) (a)

Here, m_pe is the mass of person, m_pa is the mass of parachute and a is the downward acceleration.

To obtain the net force F_net on the system, substitute 77 kg for m_pe , 5.2 kg for m_pa and

-2.5 m/s² for a in the equation F_net = (m_pe+m_pa) (a),

 F_net = (m_pe+m_pa) (a) = (77 kg + 5.2 kg) (-2.5 m/s²)

   = (-210 kg,m/s²) (1 N/1 kg,m/s²) = -210 N

W =  (m_pe+m_pa) (g)

To obtain the weight W of the system, substitute 77 kg for m_pe , 5.2 kg for m_pa and 9.81 m/s² for free fall acceleration g in the equation W =  (m_pe+m_pa) (g),

W =  (m_pe+m_pa) (g) = (77 kg + 5.2 kg) (9.81 m/s²)

    = (810 kg,m/s²) (1 N/1 kg,m/s²) = 810 N

P = F_net +W = (-210 N)+ (810 N) = 600 N

From the above observation we conclude that, the upward force exerted on the parachute by the air would be 600 N.

F_net = m_pa a

Here, m_pa is the mass of parachute and a is the downward acceleration.

To obtain the net force F_net on the parachute, substitute  5.2 kg for m_pa and -2.5 m/s² for a in the equation F_net = m_pa a,

F_net = m_pa a = (5.2 kg)(-2.5 m/s²) 

= (-13 kg.m/s²) (1 N/1 kg,m/s²) = -13 N

W =  (m_pa) (g)

To obtain the weight W of the system, substitute 5.2 kg for m_pa and 9.81 m/s² for free fall acceleration g in the equation W =  (m_pa) (g),

W =  (m_pa) (g)

= ( 5.2 kg) (9.81 m/s²)= (51 kg,m/s²) (1 N/1 kg,m/s²)

   = 51 N

D = - F_net-W+P

    = -(-13 N)-(51 N)+(600 N) = 560 N

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Problem 3:-

A 1400-kg jet engine is fastened to the fuselage of a passenger jet by just three bolts (this is the usual practice). Assume that each bolt supports one-third of the load. (a) Calculate the force on each bolt as the palne waits in line for clearance to take off. (b) During flight, the palne encounters turbulence, which suddenly imparts an upward vertical acceleration of 2.60 m/s² to the palne. Calculate the force on each bolt now. why are only three bolts used? See below figure.

Concept:-

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

So, F = ma  

W = mg

(a) First we have to find out the weight W of the engine.

To obtain the weight of the engine W, substitute 1400 kg for mass m and 9.81 m/s² for g in the equation W = mg,

W = mg = (1400 kg) (9.81 m/s²)

   = 1.37×10⁴ kg.m/s² = (1.37×10⁴ kg.m/s²) (1 N/1 kg.m/s²) = 1.37×10⁴ N

As each bolt supports 1/3 of this force, thus the force F on a bolt will be,

F = 1.37×10⁴ N/3 = 4600 N

From the above observation we conclude that, the force on each bolt would be 4600 N.

To obtain the net force F_net on the engine, substitute 1400 kg for mass of the jet engine m and 2.60 m/s² for acceleration a in the equation F_net = ma,

F_net = ma =(1400 kg) (2.60 m/s²) =3.64×10³ kg.m/s²

      =(3.64×10³ kg.m/s²) (1 N/1 kg.m/s²) =3.64×10³ N

F_up = F_net + W

To obtain the upward force F_up from the bolts, substitute 3.64×10³ N for F_net and  1.37×10⁴ N for W in the equation F_up = F_net + W,

F_up = F_net + W = (3.64×10³ N) + (1.37×10⁴ N) = 1.73×10⁴ N

So, f = F_up/3.

To obtain the force per bolt f, substitute 1.73×10⁴ N for F_up in the equation f = F_up/3,

f = F_up/3 = (1.73×10⁴ N)/3

  = 5800 N

From the above observation we conclude that, the force on each bolt would be 5800 N.

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Problem 4:-

A research ballon of total mass M is descending vertically with downward acceleration a as shown in below figure. How much ballast must be thrown from the car to give the ballon an upward acceleration a, assuming that the upward lift of the air on the ballon does not change?

Solution:-

Let us consider initially mass of the system (balloon) is, M.

So the force (f) acting on the system having downward acceleration, a will be,

f = -Ma        …… (1)

And the weight of the system (W) will be,

W = Mg       …… (2)

Where, g is the free fall acceleration of the system.

Therefore the upward force acting on the system (F) will be,

F = W + f = Mg+(-Ma)

F = Mg-Ma        …… (3)

Now mass of the system is, M-m.

So the force (f₁) acting on the system having upward acceleration, a and mass, M-m will be,

f₁ = (M-m) a      …… (4)

And the weight of the system (W₁) having mass M-m will be,

W₁ = (M-m) g       …… (5)

Where, g is the free fall acceleration of the system.

So the equation will be,

f₁+W₁ = F      …… (6)

(M-m) a + (M-m) g = Mg-Ma

Ma-ma +Mg-mg = Mg-Ma 

 ma + mg = Ma+ Mg –Mg + Ma

m(a+g) = 2Ma

m = 2Ma/(a+g)       …… (7)

From equation (7) we observed that, 2Ma/(a+g) mass of ballast must be thrown from the car to give the balloon an upward acceleration a.

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Problem 5:-

A child’s toy consists of three cars that are pulled in tandem on small frictionless rollers as shown in below figure. The cars have masses m₁ = 3.1 kg, m₂ = 2.4 kg, and m₃ = 1.2 kg. If they are pulled to the right with a horizontal force P = 6.5 N, find (a) the acceleration of the system, (b) the force exerted by the second car on the third car, and (c) the force exerted by the first car on the second car.

Solution:-

Given Data:

Mass of the first car, m₁ = 3.1 kg

Mass of the second car, m₂ = 2.4 kg

Mass of the third car, m₃ = 1.2 kg

Horizontal force on the car, P = 6.5 N

Acceleration a body is equal to the force exerted on the body divided by mass of the body.

(a) The acceleration of the system is equal to the total horizontal force acting on the system divided by total mass of the system.

So, a = F/(m₁+ m₂+ m₂) =   (6.5 N)/(3.1 kg+2.4 kg+1.2 kg)

    = (6.5 N)/(6.7 kg) = (0.97 N/kg) (1 kg. m/s²/1N) = 0.97 m/s²

Therefore the acceleration of the system would be 0.97 m/s².  

   = (1.164 kg. m/s²) (1 N/ 1 kg. m/s²) = 1.164 N

Rounding off to two significant figures, the force exerted by the second car on the third car would be 1.2 N.

(c) Force exerted by the first car on the second car would  be,

F₃₂ = (sum of mass of second car and third car) (total acceleration of the syste) = (m₂+ m₃) a

= (2.4 kg +1.2 kg) (0.97 m/s²)= (3.492 kg. m/s²) (1 N/ 1 kg. m/s²) = 3.492 N

Rounding off to two significant figures, the force exerted by the first car on the second car would be 3.5 N.

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Problem 6:-

A landing craft approaches the surface of Callisto, one of the satellite (moons) of the planet Jupiter as shown in the below figure. If an upward thrtust of 3260 N is supplied by the rocket engine, the craft descends with constant speed. Callisto has no atomsphere. If the upward thrust is 2200 N, the craft accelerates downward at 0.390 m/s². (a) What is the weight of the landing craft in the vicinity of Callisto’s surface? (b) What is the mass of the craft? (c) What is the acceleration due to gravity near the surface of Callisto?

Concept:-

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

So, F = ma      

So mass m of the body will be,

m = F/a

W = mg

So, g = W/m    

(a) As the craft descends with constant speed, so the net force will be equal to zero. Thus the thrust balances weight. As the upward thrust of 3260 N is supplied by the rocket engine, therefore the weight of the landing craft in the vicinity of Callisto’s surface will be 3260 N.

So the net force F will be,

F = 2200 N – 3260 N = -1060 N

To obtain the mass m of the craft, substitute -1060 N for F and -0.390 m/s² for acceleration a (negative sign due to downward direction) in the equation m = F/a,

m = F/a = (-1060 N)/(0.390 m/s²) = (2720 N/(m/s²)) (1kg.m/s² /1 N) = 2720 kg

From the above observation we conclude that, the mass m of the craft will be 2720 kg.

g = W/m = 3260 N/2720 kg = (1.20 N/kg) (1 kg.m/s² /1 N) = 1.20 m/s²

From the above observation we conclude that, the acceleration due to gravity g near the surface of Callisto would be 1.20 m/s².

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Problem 7:-

Two blocks are in contact on a frictionless table. A horizonatl force is applied to one block, as shown in the below figure. (a)If m₁ = 2.3 kg, m₂ = 1.2 kg, and F = 3.2 N, find the force of contact between the two blocks. (b) Show that If the same force F is applied to m₂ rather than to m₁, the force of contact between the blocks is 2.1 N, which is not the same value derived in (a). Explain.

Concept:-

Force acting on the body (F) is equal to the product of mas of the body (m) and acceleration of the body (a).So, F = ma  

From equation F = ma, the acceleration (a) of the body would be,

a = F/m   

Solution:-

(a) The acceleration a of the two block will be,

a = F/(m₁+m₂)

Here F is the horizontal force, m₁ is the mass of the block 1 and m₂ is the mass of the block 2.

The net force F_net on block 2 is from the force of contact, and will be,

F_net = m₂a

       = F m₂/(m₁+m₂)     (Since, a = F/(m₁+m₂))

F_net = F m₂/(m₁+m₂)

       = (3.2 N) (1.2 kg)/(2.3 kg+1.2 kg) = 1.1 N

From the above observation we conclude that, the force F_net of contact between the two blocks would be 1.1 N.

a = F/(m₁+m₂)

Here F is the horizontal force, m₁ is the mass of the block 1 and m₂ is the mass of the block 2.

The net force F_net on block 1 is from the force of contact, and will be,

F_net = m₁a

       = F m₁/(m₁+m₂)     (Since, a = F/(m₁+m₂))

F_net = F m₁/(m₁+m₂)

       = (3.2 N) (2.3 kg)/(2.3 kg+1.2 kg) = 2.1 N

From the above observation we conclude that, if the same force F is applied to m₂ rather than to m₁, the force of contact between the blocks would be 2.1 N.

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Problem 8:-

A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m, as shown in the below figure. A horizontal force  is applied to one end of the rope. Assuiming that the sag in the rope is negligible, find (a) the acceleration of rope and block, and (b) the force that the rope exerts on the block.

Concept:-

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

So, F = ma           

From equation F = ma, the acceleration (a) of the body would be,

a = F/m                 

(a) Treat the system as including both the block and the rope.

Thus the mass of the system will be M+m.

Here M is the mass of the block and m is the mass of the rope.

As there is one horizontal force which is acting to one end of the rope, so,

        

In accordance to Newtons second law, the horizontal force P will be,

P = (M+m) a_x

Here a_x is the horizontal acceleration of the rope and block.

From equation P = (M+m)a_x, the horizontal acceleration a_x of the rope and block will be,

a_x = P/(M+m) 

From the above observation we conclude that, the horizontal acceleration a_x of the rope and block would be P/(M+m).

Thus the net force on the block will be,

 

In this case Newton’s second law would be written as,

R = Ma_x 

Here the horizontal acceleration a_x of the block is equal to the acceleration of the block and rope system and M is the mass of the block.

To obtain the force R that the rope exerts on the block, substitute P/(M+m) for acceleration  a_x in the equation  R = Ma_x we get,

R = Ma_x= m [P/(M+m)] = [M/(M+m)] P

From the above observation we conclude that, the force R that the rope exerts on the block would be [M/(M+m)] P.

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Problem 9:-

A light beam from a satellite-carried laser strikes an object ejected from an accidently launched ballistic missile; see below figure. The beam exerts a force of 2.7×10^-5 N on the target. If the “dwell time” of the beam on the target is 2.4 s, by how much is the object displaced if it is (a) a 280-kg warhead and (b) a 2.1-kg decoy? (These displacements can be measured by observing the reflected beam.)

Concept:-

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

So, F = ma  

From the above equation F = ma, the acceleration a will be, 

a = F/m

y = ut + ½ at²    

Here u is the initial velocity, t is the time, and a is the acceleration.

Since the object ejected from an accidentally launched ballistic missile, so the initial velocity u will be equal to zero.

So, u = 0

Substitute 0 for u in the equation y = ut + ½ at²,

y = ut + ½ at² = 0×t +½ at² = ½ at²

(a) To find the displacement y which is the object displaced, first we have to find out the acceleration a.

To obtain the acceleration a, substitute 2.7×10^-5 N for F and 280 kg for m in the equation a = F/m,

a = F/m

   = (2.7×10^-5 N)/280 kg = (9.64×10^-8 N/kg) (1 kg.m/s² /1 N) = 9.64×10^-8 m/s²

y = ½ at²

   = ½ (9.64×10^-8 m/s²) (2.4 s)² = 2.8×10^-7 m

From the above observation we conclude that, the displacement y which is the object displaced from the original trajectory would be 2.8×10^-7 m.

To obtain the acceleration a, substitute 2.7×10^-5 N for F and 2.1 kg for m in the equation a = F/m,

a = F/m

   = (2.7×10^-5 N)/2.1 kg = (1.3×10^-5 N/kg) (1 kg.m/s² /1 N) = 1.3×10^-5 m/s²

y = ½ at²

   = ½ (1.3×10^-5 m/s²) (2.4 s)² = 3.7×10^-5 m

From the above observation we conclude that, the displacement y which is the object displaced from the original trajectory would 3.7×10^-5 m.

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Problem 10:-

What strength fishing line is needed to stop a 19-lb salmon swimming at 9.2 ft/s in a distance of 4.5 in.?

Concept:-

We have to calculate the force acting on the salmon.

Force (F) acting on the body is equal to the mass of the body (m) times acceleration (a) of the body.

F = ma

In terms of weight (W = mg) the above equation (F = ma) will be,

F = ma

  = (ma) (g)/g = (mg)a/g = Wa/g

Initial speed (v_i) of the salmon is, v_i = 9.2 ft/s

Final speed (v_f) of the salmon is, v_f = 0 ft/s

So the average speed of the salmon will be, v_av = (v_i + v_f) /2

                                                                             = (0 ft/s+9.2 ft/s)/2 =    = 4.6 ft/s

Time required (t) to stop the salmon is equal to the distance travelled (x) by the salmon divided by the average velocity (v_av) of the salmon.

So, t = x/ v_av

To obtain the time required (t) to stop the salmon, substitute 4.5 in for x and 4.6 ft/s for v_av in the equation  t = x/ v_av,

t = x/ v_av

  =(4.5 in)/(4.6 ft/s) =(4.5 in×0.0833 ft/1 in)/(4.6 ft/s) = (0.38 ft) /(4.6 ft/s) = 8.3×10^-2 s

To find out the   deceleration a of the salmon, substitute 9.2 ft/s for Δv and 8.3×10^-2 s for Δt in the equation a = Δv/Δt,

a = Δv/Δt

  = (9.2 ft/s)/(8.3×10^-2 s) =110 ft/s²

To find out the force on the salmon, substitute 19-lb for W, 110 ft/s² for a, and 32 ft/s² for g in the equation F = Wa/g,

F = Wa/g

   = (19-lb) (110 ft/s²)/ (32 ft/s²) = 65.31 lb

Rounding off to two significant figures, the force will be 65-lb.

From the above observation we conclude that, the force required to stop the 19-lb salmon swimming at 9.2 ft/s in a distance of 4.5 in would be 65-lb.

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Problem 11:-

A jet plane starts from rest on the runway and accelerates for takeoff at 2.30 m/s². It has two jet engines, each of which exerts a thrust of 1.40 ×10⁵ N. What is the weight of the plane?

Solution:-

Given Data:

At takeoff time, acceleration of the jet plane, a_x = 2.30 m/s²

Thrust of each jet engine, F_x = 1.40×10⁵ N

Jet plane consist two jet engines. So the total force exerts by the get plane will be,

∑ F_x = 2(1.40×10⁵ N)

        = 2.80×10⁵ N       …… (1)

Force exerted on a body is equal to the mass of the body times acceleration of the body.

So, ∑ F_x = ma_x        …… (2)

So from equation (2),

m = ∑ F_x /a_x     …… (3)

Putting the value of ∑ F_x and a_x in equation (3), mass of the jet plane will be,

m = ∑ F_x /a_x    

    = 2.80×10⁵ N/2.30 m/s² = (1.22 ×10⁵ N/m/s²) (1 kg. m/s² / 1 N) = 1.22 ×10⁵ kg     ……. (4)

W = mg   …… (5)

We know that that, the value of acceleration due to gravity (g) on the surface of earth is 9.81 m/s².

So putting the value of m and g in equation (5), weight of the jet plane would be,

W = mg  

    = (1.22×10⁵ kg) (9.81 m/s²) = (11.9682×10⁵ kg. m/s²) (1 N/1 kg. m/s²)

    = 1.19682×10⁶ N     …… (6)

Rounding off to three significant figures, weight of the jet plane would be 1.20×10⁶ N.

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