AP Physics B & C – Few More Multiple Choice Practice Questions on Friction

"What you are will show in what you do."

– Thomas A. Edison

Questions involving friction were discussed on this site earlier. You can access them by clicking on the label ‘friction’ below this post or by trying a search for ‘friction’, using the search box provided on this page.

Today we will discuss some more multiple choice practice questions on friction:

(1) A uniform chain is placed on a rough horizontal table so that one end of the chain hangs down over the edge of the table. When 20% of the length of the chain hangs over the edge, it starts sliding. What is the coefficient of static friction between the chain and the table?

(a) 0.1

(b) 0.15

(c) 0.25

(d) 0.35

(e) 0.5

When the chain just begins to slide, the frictional force between the chain and the table is maximum and is called limiting frictional force which is equal to μ_sN where μ_s is the coefficient of static friction between the chain and the table and N is the normal force.

Here N = 0.8mg and μ_sN = 0.2mg where m is the mass of the chain.

[Note that the weight of 20% of the length of the chain (= 0.2mg) balances the frictional force μ_sN].

Therefore, μ_s×0.8 mg = 0.2 mg from which μ_s = 0.25

(2) A wooden block is placed on a horizontal surface and a horizontal force equal to the limiting frictional force is applied on it. If the coefficient of static friction and the coefficient of kinetic friction are respectively 0.5 and 0.4 and the acceleration due to gravity is 10 ms^–2, the acceleration of the wooden block is

(a) 1 ms^–2

(b) 1.1 ms^–2

(c) 1.4 ms^–2

(d) 1.5 ms^–2

(e) zero

The limiting (or, maximum) frictional force is μ_s mg where μ_s is the coefficient of static friction, and mg is the weight of the wooden block. When the block moves along the horizontal surface, the frictional force acting is reduced to kinetic frictional force and is equal to μ_k mg where μ_k is the coefficient of kinetic friction.

[If the frictional force were not reduced, the block would have moved with uniform velocity since there would be no net force].

The net force acting on the block is μ_s mg – μ_k mg and the acceleration a of the block is given by,

a = (μ_s mg – μ_k mg)/m = (μ_s – μ_k) g = (0.5 – 0.4) × 10 = 1 ms^–2

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(3) A slab of mass 50 kg is placed on frictionless horizontal floor and a block of mass 10 kg is placed on the slab as shown. The coefficient of static friction and the coefficient of kinetic friction between the slab and the block are respectively 0.4 and 0.2 and the acceleration due to gravity is 10 ms^–2. A horizontal force equal to 30 N acts on the block. The acceleration of the slab will be

(a) 0.1 ms^–2

(b) 0.2 ms^–2

(c) 0.3 ms^–2

(d) 0.5 ms^–2

(e) 1.2 ms^–2

The limiting frictional force (μ_sN) between the block and the slab is given by

μ_sN = μ_sMg = 0.4×10×10 = 40 N

The block will not slide along the slab since the applied force (= 30 N) is less than the limiting frictional force. So the slab and the block will move together, with acceleration a given by

a = 30 N/(50+10)kg = 0.5 ms^–2.

[Since there is no relative motion between the slab and the block, the coefficient of kinetic friction does not play any role in the above problem. It just serves the purpose of a distraction].

The above questions may be ‘enjoyed’ by AP Physics B & C aspirants. The following questions are specifically for AP Physics C aspirants:

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(4) A slab A of mass M = 6 kg is placed on rough horizontal floor. The coefficient of kinetic friction between the floor and the slab is 0.1. A block B of mass m = 4 kg is placed on the slab (Fig.). The coefficient of static friction between the slab and the block is 0.2. When a horizontal force of magnitude F is applied on the slab, the block B just begins to slide along the slab A. What is the value of F? (Take g = 10 ms^–2).

(a) 40 N

(b) 30 N

(c) 15 N

(d) 10 N

(e) 8 N

The limiting frictional force between the slab and the block is μ_smg where μ_s is the coefficient of static friction between the slab and the block.

When the force F is applied on the slab, the entire system containing the slab and the block must move with an acceleration a so that the inertial force (= ma) on the block just balances the frictional force.

[The inertial force must be infinitesimally greater than the frictional force for the slipping to occur].

Therefore, we have μ_smg = ma so that a = μ_sg = 0.2×10 = 2 ms^–2

If the floor were smooth, the force to be applied on the slab to attain the slipping condition would have been a(M+m).

But since the floor is rough, the applied force has to overcome the force of kinetic friction [μ_k(M+m)g] between the floor and the slab.

Therefore, F = μ_k(M+m)g + a(M+m)

Thus F = (M+m)( μ_kg + a) = 10(1+2) = 30 N.

[The above question could be made a little more difficult if the coefficient of static friction (say, 0.25) between the floor and the slab also is given (to distract you). You should remember that once the slab slides along the floor, the friction called into play is kinetic friction and hence your answer will be unchanged].

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(5) A man of mass M = 80 kg tries to pull down a tree using a rope. He exerts the pulling force at an angle of 30º with the horizontal (Fig.). If the coefficient of static friction between the man’s feet and the ground is 0.6 and the gravitational acceleration is 10 ms^–2, the maximum pulling force he can exert (before he slips) is approximately

(a) 50 N

(b) 100 N

(c) 200 N

(d) 400N

(e) 600 N

If the man exerts too much pulling force his feet will slip and hence the maximum pulling force is determined by the coefficient of friction and the angle of pull. If T is the maximum tension produced in the rope, the maximum pulling force, F = T.

The vertical component of T which acts upwards reduces the normal force N and we have

N = Mg – T sin 30º = 80×10 – T/2

The maximum frictional force = μ_sN = 0.6×(800 – T/2).

The horizontal component of the maximum tension T = T cos 30º = T×(√3)/2

The frictional force should balance the horizontal component of tension so as to prevent the man from slipping.

Therefore we have

0.6×(800 – T/2) = T ×(√3)/2

Or, 960 – 0.6 T = 1.732 T so that T = 960/2.332 = 400 N, approximately

Multiple Choice Practice Questions on Vectors (for AP Physics B & C)

There is no substitute for hard work.

– Thomas A. Edison

AP Physics C aspirants are expected to have a fairly good idea regarding vector methods in Physics. AP Physics B aspirants also should have some idea of vectors. Today we will discuss some multiple choice practice questions on vectors.

(1) The magnetic flux density B at a point P is 0.5 tesla. What is the maximum number of components into which the vector B can be resolved?

(a) 1

(b) 2

(c) 3

(d) 6

(e) infinite

The resultant of all possible components should make the vector B of the given direction and of magnitude 0.5 tesla. We can imagine an infinite number of components in various directions to combine and produce the field B and hence the correct option is (e).

(2) The force F acting on a particle has its line of action (direction) lying in the XY plane and is inclined at an angle θ with the x-axis. What is the z-component of the force F?

(a) zero

(b) F cos θ

(c) F sin θ

(d) F/cos θ

(e) F/sin θ

Since the direction of the vector F is lying in the XY plane it has no z-component. The correct option is (a).

(3) The resultant of two forces F₁ and F₂ of the same magnitude F has magnitude F itself. The angle between F₁ and F₂ is

(a) 30º

(b) 45º

(c) 60º

(d) 90º

(e) 120º

The magnitude F of the resultant force is given by

F = √(F₁² + F₂² + 2 F₁ F₂ cos θ) where θ is the angle between the forces F₁ and F₂.

Since the magnitudes F₁ and F₂ of the two forces are the same and equal to F, we have

F = √(F² + F² + 2 F F cos θ)

Squaring, F² = F² + F² + 2 F² cos θ

This gives cos θ = – ½ so that θ = 120º

[The adjoining figure will be useful to understand how the two forces produce the resultant satisfying the given conditions].

(4) The resultant of two forces is 40 N and the smaller force, which has magnitude 30 N, is normal to the resultant. The larger force is

(a) 40 N

(b) 45 N

(c) 50 N

(d) 60 N

(e) 80 N

The forces and their resultant are shown in the adjoining figure.

Since the resultant R of magnitude 40 N is perpendicular to the force F₁ of 30 N, we have (from the right angled triangle),

F₂ = √(30² + 40²) = 50 N.

(5) Rain drops are falling vertically with a speed of 10 ms^–1. A boy holding an umbrella runs southward with a speed of 5 ms^–1. What is the direction in which he should hold his umbrella so that he will not get drenched?

(a) At an angle of tan^–1(1/2) with the vertical, towards north

(b) At an angle of tan^–1(1/2) with the vertical, towards south

(c) At an angle of tan^–1(2) with the vertical, towards south

(d) At an angle of 30º with the vertical, towards north

(e) At an angle of 30º with the vertical, towards south

In the figure the velocities of the boy and rain drops are represented respectively by the vectors OA and OB.

The relative velocity of the rain drops with respect to the boy = (velocity of rain drops) – (velocity of boy).

The quantity on the right hand side is the vector sum of the velocity of rain drops and the negative of the velocity (reversed velocity shown by vector OC) of the boy. This is given by the vector OD which makes an angle θ with the vertical and is given by

tan θ = 5/10 = ½.

Therefore, the boy should hold his umbrella at an angle of tan^–1(1/2) with the vertical, towards south.

The following multiple choice questions are specifically meant for AP Physics C aspirants:

(6) Suppose that the ground coincides with the XZ plane of a right handed Cartesian coordinate system and i, j, k are unit vectors along x, y, z directions respectively. An inclined plane of inclination 60º is placed on the ground and a force F = – 20 j N is applied at a point on the inclined plane. What is the component of this force parallel to the inclined plane?

(a) 20/√3 N

(b) 10 N

(c) 10√3 N

(d) 5√3 N

(e) Zero

The force F of magnitude 20 N is directed vertically downwards (along the negative y-direction).

With reference to the figure, you can easily see that the component of F parallel to the inclined plane is 20 sin 60º = 10√3 N.

(7) The magnitude of the area of a parallelogram formed by the vectors A = i + 2 j + 2 k metre and B = 2 i – 2 j + k metre as adjacent sides is (i, j, k are unit vectors)

(a) √78 m²

(b) 9 m²

(c) √91 m²

(d) 11 m²

(e) 12 m²

The magnitude of the area of the parallelogram is equal to the magnitude of the vector product (cross product) of the vectors A and B.

[Remember that area is a vector].

Area = A × B = (i + 2 j + 2 k) × (2 i – 2 j + k)

Or, Area = – 2 k – j – 4 k + 2i + 4 j + 4 i

= 6 i + 3 j – 6 k

The magnitude of area vector = √(6² + 3² + 6²) = √81 = 9 m²

You will find similar useful questions (with solution) on vectors here.

AP Physics B & C - Answer to Free Response Practice Question on Kinematics

In the post dated August 31, 2010 a free response practice question on kinematics was given to you. As promised, I give below a model answer for the same along with the question:

A ball at a height H is released from rest at the instant t = 0 and it strikes a fixed touch plate P, inclined at 45º with respect to the vertical, at the instant t = t₁. The point where the ball strikes the plate is at a height h (fig.). The collision between the ball and the plate is elastic. After the collision the ball continues its motion and strikes the ground at the instant t = t₂. Now, answer the following questions:

(a) If the height h where the ball strikes the plate is varied, how will you indicate, graphically, the variation of t₁ with (H – h)? Use the coordinate axes as shown in the figure below to show the nature of variation.

(b) After the collision what is the nature of the path followed by the ball? Pick out your answer from the following three options by putting a tick (√) mark against the correct one:

Straight line path ……

Parabolic path ……

Arc of a circle …..

Justify your answer.

(c) Derive an expression for the total time of flight t₂ of the ball.

(d) Determine the horizontal displacement of the ball when it strikes the ground.

AP Physics C aspirants should answer the following part also:

(e) For a given value of H determine the value of h for which the time of flight t₂ of the ball is maximum.

(a) The nature of variation of the displacement H – h suffered by the ball in time t₁ is shown in the adjoining figure.

[The graph is obtained from the equation H – h = 0 + ½ gt₁² and is parabolic]

(b) Just after the collision, the ball moves horizontally. Its horizontal velocity component remains unchanged and it picks up vertical velocity (downward) because of the gravitational pull. The path followed is therefore parabolic as in the case of any projectile having two dimensional motion.

(c) During the time t₁ the ball suffers a vertical displacement (H – h) and hence we have

H – h = 0 + (½) g t₁²

This gives t₁ = [2(H – h)/g]^1/2

Since the touch plate is inclined at 45º with the vertical and the collision is elastic, the ball moves horizontally just after the collision. Its horizontal velocity remains constant throughout its trajectory. But it has vertical acceleration g directed downwards and the time t’ taken for the vertical displacement h is given by

h = 0 + (½) g t’²

[Note that the initial vertical velocity is zero just after the hit at the touch plate].

This gives t’ = [2h/g]^1/2

The total time of flight t₂ = t₁ + t’ = [2(H – h)/g]^1/2 + [2h/g]^1/2

Or, t₂ = [2/g]^1/2 [(H – h)^1/2 + h^1/2]

(d) The horizontal speed v of the ball after the collision at the touch plate is the same as the vertical speed just when the ball hits the touch plate. Therefore we have

v = [2g(H – h)]^1/2

[This follows from v² = 0 + 2g(H – h)]

The horizontal displacement R of the ball is given by

R = vt’ = [2g(H – h)]^1/2[2h/g]^1/2 = 2[h(H – h)]^1/2

[You might have come across this equation earlier in fluid mechanics. This is the horizontal range of a liquid jet coming out through a side hole in a liquid tank (See the post dated 3^rd December 2007)].

(e) For a given height H the time of flight t₂ will change when the height h of the touch plate is changed. the time of flight t₂ will be maximum when dt₂/dh = 0

Therefore, [2/g]^1/2[–(½)(H – h)^–1/2 + (½)h^–1/2] = 0

This gives H – h = h from which h = H/2