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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

So the mass m of the object would be,

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 vav will be,

t = x/ vav

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

a = Δv/Δt

Solution:-

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

t = x/ vav = (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

To find out the net force F on the plane, substitute 52,000 lb for W, 130 ft/s2 for a and 32 ft/s2 for free fall acceleration g in the equation F =(W/g) (a),

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

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

So, Fc = F-T

To obtain the force exerted by the catapult Fc, substitute 2.1×105 lb for F and 24,000 lb for T in the equation Fc = F-T,

Fc = 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 Fc 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

Solution:-

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

So, Fnet = (mpe+mpa) (a)

Here, mpe is the mass of person, mpa is the mass of parachute and a is the downward acceleration.

To obtain the net force Fnet on the system, substitute 77 kg for mpe , 5.2 kg for mpa and

-2.5 m/s2 for a in the equation Fnet = (mpe+mpa) (a),

Fnet = (mpe+mpa) (a) = (77 kg + 5.2 kg) (-2.5 m/s2)

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

The weight W of the system will be,

W =  (mpe+mpa) (g)

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

W =  (mpe+mpa) (g) = (77 kg + 5.2 kg) (9.81 m/s2)

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

If P is the upward force of the air on the system (parachute) then,

P = Fnet +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.

(b) The net force Fnet on the parachute will be,

Fnet = mpa a

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

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

Fnet = mpa a = (5.2 kg)(-2.5 m/s2)

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

The weight W of the parachute will be,

W =  (mpa) (g)

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

W =  (mpa) (g)

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

= 51 N

If D is the downward force of the person on the parachute then,

D = - Fnet-W+P

= -(-13 N)-(51 N)+(600 N) = 560 N

__________________________________________________________________________________________________________________

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

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 = 32 ft/s2).

W = mg

Solution:-

(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.

(b) To find out the force f on each bolt, first we have to find out the upward force Fup on the bolt. Again to obtain the upward force Fup on the bolt, we have to obtain the net force Fnet on the engine.

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

Fnet = 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

So the upward force Fup from the bolts will be equal to the sum of net force Fnet on the engine and weight of the engine W.

Fup = Fnet + W

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

Fup = Fnet + W = (3.64×103 N) + (1.37×104 N) = 1.73×104 N

The force per bolt f will be equal to the 1/3 of the upward force Fup from the bolts.

So, f = Fup/3.

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

f = Fup/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)

Again let us consider m is the mass of single ballast which is thrown from the balloon.

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.

To give the balloon an upward acceleration a, the upward force (F) must be equal to the addition of the force (f1) acting on the system having upward acceleration, a and mass, M-m and the weight of the system (W1) having mass M-m will be.

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.

(b) Force exerted by the second car on the third car would  be,

F32 = (mass of third car) (total acceleration of the system) = m3 a = (1.2 kg) (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

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 = 32 ft/s2).

W = mg

So, g = W/m

Solution:-

(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.

(b) An upward thrust of 3260 N is supplied by the rocket engine; the craft descend with constant speed. If the upward thrust is 2200 N, the craft accelerates downward at 0.390 m/s2.

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.

(c) To find out the acceleration due to gravity g near the surface of Callisto, substitute 3260 N for W and 2720 kg for m in the equation g = W/m,

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.

________________________________________________________________________________________________________________

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 Fnet on block 2 is from the force of contact, and will be,

Fnet = m2a

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

To obtain the force Fnet of contact between the two blocks, substitute 3.2 N for F,1.2 kg for m2 and 2.3 kg for m1 in the equation Fnet = F m2/(m1+m2),

Fnet = 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 Fnet of contact between the two blocks would be 1.1 N.

(b) 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 Fnet on block 1 is from the force of contact, and will be,

Fnet = m1a

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

To obtain the net force Fnet on block 1, substitute 3.2 N for F,1.2 kg for m2 and 2.3 kg for m1 in the equation Fnet = F m2/(m1+m2),

Fnet = 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

Solution:-

(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) ax

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

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

ax = P/(M+m)

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

(b) Now consider only the block. The horizontal force does not act on the block, instead there is the force of the rope on the block. We will assume that, the magnitude of the force is R, and this the only relevant force on the block.

Thus the net force on the block will be,

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

R = Max

Here the horizontal acceleration ax 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  ax in the equation  R = Max we get,

R = Max= 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.

__________________________________________________________________________________________________________________

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

In accordance to equation of motion, the distance y travelled by the body will be,

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

Solution:-

(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

To find the displacement y which is the object displaced from the original trajectory, substitute 9.64×10-8 m/s2 for a and 2.4 s for t in the equation  y = ½ at2,

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.

(b) 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 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

To find the displacement y which is the object displaced from the original trajectory, substitute 1.3×10-5 m/s2 for a and 2.4 s for t in the equation  y = ½ at2,

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.

_________________________________________________________________________________________________________________

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

Solution:-

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

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

So the average speed of the salmon will be, vav = (vi + vf) /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 (vav) of the salmon.

So, t = x/ vav

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

t = x/ vav

=(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

The deceleration a of the salmon will be, a = Δv/Δt.

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.

__________________________________________________________________________________________________________________

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, ax = 2.30 m/s2

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

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

∑ Fx = 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, ∑ Fx = max        …… (2)

So from equation (2),

m = ∑ Fx /ax     …… (3)

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

m = ∑ Fx /ax

= 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)

Weight of a body is equal to the mass of the body times (m) acceleration due to gravity (g) on that surface.

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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