Physics

5.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

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Given, the mass m = 5 kg

The resultant force F of the two forces 6N and 8N is given by,

F = $\sqrt{8^{2} + 6^{2}}$ = 10 N

Acceleration is given by F = ma, a = F/m = 10/5 m/s² = 2 m/s²

$\tan ? θ$ = 6/8, $θ$ = 36.86 $°$

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5.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s^-1 to 3.5 m s^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

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Initial speed, u = 2.0 m/s, the final speed v = 3.5 m/s, the time t = 25 s

From the relation v = u + at, we get acceleration a = (v-u)/t = (3.5 – 2.0 )/ 25s = 0.06 m/s²

The force F = ma, F = 3 $*$ 0.06 N = 0.18 N

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5.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s-1. How long does the body take to stop?

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The retarding force, F = 50N, the mass m = 20 kg

From the equation F = ma, we get acceleration a = 50 / 20 m/s² = 2.5 m/s²

The initial speed, u = 15 m/s, the final speed v =0, from the relation v = u-at, we get

T = u/a = 15/2.5 s = 6s

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5.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,

(a) Just after it is dropped from the window of a stationary train

(b) Just after it is dropped from the window of a train running at a constant velocity of 36 km/h

(c) Just after it is dropped from the window of a train accelerating with 1 m s^-2

(d) Lying on the floor of a train which is accelerating with 1 m s^-2, the stone being at rest relative to the train. (Neglect air resistance throughout.)

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(a) Mass of the stone, m = 0.1 kg, Acceleration due to gravity = 10 m/s²

Net force = 0.1 $*$ 10 N = 1.0 N. The direction of force is downwards

(b) Since the train is moving at a constant velocity, the acceleration imparted by the train is zero. Only acceleration acting on the stone is gravity. So the force will be 1.0 N, acting downwards

(c) The acceleration of the train = 1 m/s² so the force exerted by the train = 0.1 $*$ 1 N = 0.1 N and it is acting in the horizontal direction. But when the stone is dropped, the force started acting on the stone is only due to gravity, the train moving force has no effect. So the net

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(a) Mass of the stone, m = 0.1 kg, Acceleration due to gravity = 10 m/s²

Net force = 0.1 $*$ 10 N = 1.0 N. The direction of force is downwards

(d) When the stone is lying on the train floor, the only force acting on the stone is due to the acceleration of the train. So the net force on the stone = 0.1 $*$ 1 N. The direction is horizontal along the motion of the train

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5.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,

(a) During its upward motion

(b) During its downward motion

(c) At the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?

(Ignore air resistance.)

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(a) During the upward motion f the pebble, the acceleration due to gravity acts downwards. The magnitude of the force is

F = MA = 0.05 kg $*$ 10 m/s² = 0.5 N and the direction of force is downwards.

(b) During its downward motion – the force is 0.5 N and the direction is downwards.

(c) When the pebble is thrown at an angle of 45 $°$ with the horizontal direction, it will have both horizontal component of force and vertical component of force during its upward journey, When it reaches the maximum height, the vertical component of velocity will becomes zero and the force acting will be 0.5 N and acting downwards.

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5.1 Give the magnitude and direction of the net force acting on

(a) A drop of rain falling down with a constant speed

(b) A cork of mass 10 g floating on water

(c) A kite skillfully held stationary in the sky

(d) A car moving with a constant velocity of 30 km/h on a rough road

(e) A high-speed electron in space far from all material objects, and free of electric and magnetic fields

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(a) Since the raindrop is falling at a constant speed, the acceleration is zero. When the acceleration is 0, the force acting on the drop will also become zero ( since F = ma)

(b) Since the cork is floating on water, the weight of the cork is balanced by upward force of water. So the net force on the cork is 0

(c) Since the kite is held stationary in the sky, the net force is 0

(d) Since the car is moving at a constant velocity, the acceleration is 0, hence the net force is also 0

(e) The net force acting on the high-speed electron will be zero since the electron is far from the material objects and free of electric and magnetic field

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12.10 A refrigerator is to maintain eatables kept inside at 9 $?$ . If room temperature is 36 $?$ , calculate the coefficient of performance.

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Temperature inside the refrigerator, $T_{1}$ = 9 $?$ = 9 + 273 K = 282 K

Room temperature, $T_{2}$ = 36 $?$ = 36 + 273 = 309 K

Coefficient of performance = $\frac{T_{1}}{T_{2} - T_{1}}$ = $\frac{282}{309 - 282}$ = 10.44

Therefore, the coefficient of performance is 10.44

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12.9 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13)

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F

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Total work done by the gas from D to E to F = Area of $? D E F$ = $\frac{1}{2}$ EF $* D F$

Where DF = Change in pressure = 600 – 300 = 300 N/ $m^{2}$

FE = change in volume = 5-2 = 3 $m^{3}$

Area of $? D E F =$ $\frac{1}{2} *$ 3 $* 300$ = 450 J

Therefore work done by the gas from D to E to F is 450 J.

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12.8 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?

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Heat is supplied to the system at a rate of 100W

Hence, heat supplied, Q = 100 J/s

The system performs at the rate of 75 J/s

Hence, work done, W = 75 J/s

From the 1st law of Thermodynamics, we have Q = U + W, where U is the internal energy

U = Q – W = 100 – 75 = 25 J/s = 25 W

Therefore the internal energy of the given electric heater increases at a rate of 25 W.

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12.7 A steam engine delivers 5.4*108J of work per minute and services 3.6 * 109J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

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