Laws of Motion

A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is μ. Let the mass of the box be m. a) at what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane? b) what is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a > θ c) what is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed? d) what is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration a.

a) As the box starts to slide down the plane, $\mu=\tan \theta$ $ \theta=\tan ^{-1}(\mu) $ b) If $a>\theta$, the angle of inclination will be the angle of repose and the net force acting will be...

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A cricket bowler releases the ball in two different ways a) giving it only horizontal velocity and b) giving it horizontal velocity and a small downward velocity. The speed vs at the time of release is the same. Both are released at a height H from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

a) $\frac{1}{2} v_{z}^{2}=g H \Rightarrow v_{z}=\sqrt{2 g H}$ Speed at ground is given as: $\sqrt{v_{s}^{2}+v_{z}^{2}}=\sqrt{v_{s}^{2}+2 g H}$ b)$\frac{1}{2} m v_{s}^{2}+m g H$ is the total energy...

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There are three forces F1, F2, and F3 acting on a body, all acting on a point P on the body. The body is found to move with uniform speed. a) show that the forces are coplanar b) show that the torque acting on the body about any point due to these three forces is zero

a) The body's acceleration is zero because the resultant force of the three forces F1, F2, and F3 on a location on the body is zero. The directions of forces F1 and F2 are in the plane of the paper,...

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Block A of weight 100N rests on a frictionless inclined plane of slope angle 30o. A flexible cord attached to A passes over a frictionless pulley and is connected to block B of weight W. Find the weight W for which the system is in equilibrium.

Equilibrium between $A$ or $B$, Then we know that, $\mathrm{mg} \sin 30^{\circ}=\mathrm{F}$ $ \begin{array}{l} 1 / 2 \mathrm{mg}=\mathrm{F} \\ \mathrm{F}=(1 / 2)(100)=50 \end{array} $ Therefore,...

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A mass of 2 kg is suspended with thread AB. Thread CD of the same type is attached to the other end of 2 kg mass. Lower thread is pulled gradually harder and harder in the downward direction so as to apply force on AB. Which of the threads will break and why?

As the mass 2 kg acts downward, the force acting on the thread AB is equal to the force F. As a result, the force exerted on the AB is 2 kg more than on the D, and the thread AB breaks.

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A body of mass 10 kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is a) 1 m/s2 at an angle of tan-1 (4/3) w.r.t 6N force b) 0.2 m/s2 at an angle of tan-1 (4/3) w.r.t 6N force c) 1 m/s2 at an angle of tan-1(3/4) w.r.t 8N force d) 0.2 m/s2 at an angle of tan-1(3/4) w.r.t 8N force

The correct answers are a) 1 m/s2 at an angle of tan-1 (4/3) w.r.t 6N force c) 1 m/s2 at an angle of tan-1(3/4) w.r.t 8N force

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One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is : (i) T (ii) T – mv2/l (iii) T + mv2/l (iv) 0 T is the tension in the string. [Choose the correct alternative].

T is the particle's net force, and it is directed towards the centre. It gives the particle the centrifugal force it needs to travel in a circle.

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A train runs along an unbanked circular track of radius of 30 m at a speed of 54 km/h. The mass of the train is 106 kg. What provides the centripetal force required for this purpose the engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

The track's radius is 30 metres. The train's speed = 54 km/h = 54 x (5/18) = 15 m/s The train's mass is 106 kg. The force of lateral friction created by the rails on the train wheels provides the...

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Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms-2. What is the net force on the man? If the coefficient of static friction, between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg.)

Here, the conveyor belt's acceleration is a = 1 ms-2. s=0.2 is the static friction coefficient. m = 65 kg m = 65 kg m = 65 kg m = 65 kg m = 65 kg Ma = 65 x 1 = 65N is the Net Force. The friction...

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A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

The stone weighs 0.25 kilogramme. r = 1.5 m Radius n= 40/60 = (23) rev/sec is the number of revolutions per second. = 2n = 2 x 3.14 x (23) is the angular velocity. The centripetal force is provided...

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Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a tight string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

Given, A body mass of 10 kilogrames (m1) B, m2 = 20 kg, 600 N horizontal force m = m1 + m2 = 30 kg is the total mass of the system. Using Newton's second rule of motion, we can calculate ma = f...

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. A man of mass 70 kg, stands on a weighing machine in a lift, which is moving (a) upwards with a uniform speed of 10 ms-1. (b) downwards with a uniform acceleration of 5 ms-2. (c) upwards with a uniform acceleration of 5 ms-2. What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

m = 70 kg 10 m/s2 = g In each scenario, the weighing machine measures the response R, or apparent weight. (a) The lift's acceleration equals 0 when it travels upwards at a uniform speed of 10 m/s. R...

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A truck starts from rest and accelerates uniformly at 2.0 ms-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)

u = 0 is the initial velocity. a = 2 ms-2, a = 2 ms-2, a = 2 ms-2, a = 2 m   t=10s t=10s t=10s t=10   We get v = u + at using the equation v = u + at.   20 m/s = v = 0 + 2 x 10  ...

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A body of mass 0.40 kg moving initially with a constant speed of 10 ms-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = -5 s, 25 s, 100 s.

Given, Body mass is 0.40 kg. u = 10 m/s initial velocity f = -8 N force (retarding force) Using the formula S = ut + (12) at2, (a) At time t = – 5 s, position From t = 0 s, the force acts on the...

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The driver of a three-wheeler moving at a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg, and the mass of the driver is 65 kg.

Given, u=36 km/h is the initial velocity. v = 0 is the final velocity. The three-mass wheeler's is m1=400 kg. The driver's mass is m2 = 65 kg. The time it took to bring the car to a complete stop...

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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 with1 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.

(a) Stone mass = 0.1 kg 10 ms^(-2)= acceleration   F = mg = 0.1 x 10 = 1.0 N is the net force.   The force is applied vertically and downwards.   (b) The train maintains a steady...

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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 Solutions change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance

(a) The acceleration due to gravity acts downwards throughout the upward motion of the pebble, thus the magnitude of the force on the pebble is 0.5 N = F = mg = 0.05 kg x 10 ms-2 The force is in a...

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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.

(a)The raindrop continues to fall at the same rate. As a result, the acceleration will be zero. Because F = ma, the force exerted on the drop will be zero when the acceleration is zero. (b) The cork...

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