MHT-CET Physics · Laws of Motion

Newton's Laws: Lifts, Pulleys, Circles and Power

The net force on a body equals its mass times its acceleration, measured in a frame that is not itself accelerating; applied body by body it gives the reading in a lift, the push between blocks, the tension in a pulley string, and — with v²/r as the acceleration — the forces in circular motion.

All Laws of Motion notes MHT-CET Physics strategy

Why this matters

25 PYQs, 3 of them HARD. Twelve apply F = ma to lifts, blocks in contact, pulleys and ropes, and ask which frames are inertial. Nine combine force with work, power or circular motion — power from a time-varying force, a conical pendulum, the string's tension at the bottom of a swing. Four are braking and penetration problems. Three cards.

Concept 1 of 3: Lifts, Blocks in Contact and Pulleys

Draw each body and add the forces along its motion. A person in a lift feels the floor's push N = m(g + a) going up with acceleration a, m(g − a) going down; a spring balance reads the same way. Blocks pushed together share one acceleration F/(total mass), and the push on the far block is its own mass times that acceleration. In an Atwood machine, the net pull is the difference of the weights and the mass to accelerate is the total. Only frames moving at constant velocity are inertial: an accelerating train, a merry-go-round and a plane taking off are not. Gravity acts without contact; friction, normal reaction and viscosity need it.

Definition

$\sum F = ma$ . Lift: $N = m(g \pm a)$ (+ accelerating up).
Down at g/3 reads 20 N ⇒ up at g/3 reads 40 N; stationary : down = 4 : 3 ⇒ a = g/4.
Blocks in contact: $a = \dfrac{F}{m_1 + m_2}$ , force on the far block $m_2a$ (5 N on 6 + 4 kg ⇒ 2 N).
Atwood with a rider m on one of two masses M: $a = \dfrac{mg}{2M + m}$ .
Cable of a lift accelerating up: $T = m(g + a)$ . Same force on two masses: $a = \dfrac{A_1A_2}{A_1 + A_2}$ .

Second law

\sum \vec{F} = m\vec{a}, \qquad N_{\text{lift}} = m(g \pm a)

Worked example

Masses of 3 kg and 5 kg hang over a smooth pulley. Acceleration and string tension? (g = 10 m/s²)

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q12Moderate

Example 1 · Laws of Motion · Newton's Laws — Force, Tension, Lift, and Connected Blocks

Two identical blocks each of mass

M

attached to the ends of a massless inextensible string which passes over a pulley with a fixed axis as shown below. A small mass

m

is now placed on the block B. The acceleration with which the two blocks move together is [

g =

gravitational acceleration]

Subtracting a for a lift accelerating up

Accelerating UPWARD (starting up, or slowing on the way down) the floor must push harder: N = m(g + a). The reading falls only when the acceleration points down.

Using the applied force for the far block

The push on the far block is only what accelerates it: m₂F/(m₁ + m₂). 5 N on a 6 kg and a 4 kg block gives 2 N on the 4 kg block.

Drill 11 more on lifts, blocks in contact and pulleys

Concept 2 of 3: Force, Work, Power and Circular Dynamics

With a vector force, work is F·r and power F·v; a time-varying force gives v by integrating F/m, then P = F·v at the instant asked. A constant force from rest gives a = F/m and v = at. In circular motion the net inward force is mv²/r: for a car on a flat road friction supplies it, so v_max = √(μrg); for a conical pendulum it is mg tan θ = mgr/√(L² − r²); for a bob released from horizontal, v² = 2gL at the bottom and T = mg + mv²/L = 3mg. A gun firing n bullets a second pushes back with n m v.

Definition

$W = \vec{F}\cdot\vec{r}$ , $P = \vec{F}\cdot\vec{v}$ (F = tî + 2t²ĵ, 1 kg, t = 3 s ⇒ 337.5 W).
Flat curve: $v_{\max} = \sqrt{\mu rg}$ (half the speed ⇒ μ/4).
Conical pendulum: $F_c = \dfrac{mgr}{\sqrt{L^2 - r^2}}$ .
Released from horizontal: $T_{\text{bottom}} = 3mg$ . Vertical circle: $T_{\max} - T_{\min} = 6mg$ .
Gun recoil: $F = n\,mv$ (30 g at 1000 m/s, 300 N ⇒ 10 per second).

Work and power

W = \vec{F}\cdot\vec{r}, \qquad P = \vec{F}\cdot\vec{v}, \qquad F_c = \frac{mv^2}{r}

Worked example

A 2 kg body starts from rest under a force (4î + 2ĵ) N. Its speed after 5 s, and the power then?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q50Moderate

Example 2 · Laws of Motion · Newton's Laws — Force, Tension, Lift, and Connected Blocks

The string of a pendulum of length 'L' is displaced through

90^\circ

from the vertical and released. Then the maximum strength of the string in order to withstand the tension, as the pendulum passes through the mean position is (

m =

mass of pendulum,

g =

acceleration due to gravity)

Taking tension at the bottom as mg

At the bottom the string must also supply the centripetal force: T = mg + mv²/L. Released from horizontal, that is 3mg.

Using average power for power at an instant

Power at an instant is F·v, and it grows as the body speeds up. A 1 kg body pushed from rest to 10 m/s in 2 s receives 25 W at t = 1 s but 50 W at t = 2 s.

Drill 8 more on force, work, power and circular dynamics

Concept 3 of 3: Braking, Penetration and Avoiding Collision

A constant retarding force gives a constant deceleration, so v² = u² − 2as works. A bullet that halves its speed in 30 cm has lost three quarters of its kinetic energy there; the remaining quarter needs a third as much distance, 10 cm more. A faster car braking behind a slower one avoids collision if their relative speed reaches zero within the gap: s ≥ (v_A − v_B)²/2a. On a velocity–time graph the distance in any interval is the area under it.

Definition

$v^2 = u^2 - 2as$ ; penetration: V → V/2 in 30 cm ⇒ 10 cm more to stop.
No collision: $s \geq \dfrac{(v_A - v_B)^2}{2a}$ .
Distance from a v–t graph = area (last 2 s of a trapezium profile ⇒ 1/4 of the total).

Uniform retardation

v^2 = u^2 - 2as, \qquad s_{\text{safe}} = \frac{(v_A - v_B)^2}{2a}

Worked example

A bullet at 200 m/s slows to 100 m/s after 6 cm of wood. How much further does it go?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 1 · Q31Moderate

Example 3 · Laws of Motion · Newton's Laws — Force, Tension, Lift, and Connected Blocks

A bullet is fired on a target with velocity V. Its velocity decreases from V to V/2 when it penetrates 30 cm in a target. Through what thickness it will penetrate further in the target before coming to rest?

Assuming speed falls linearly with distance

Speed squared falls linearly with distance. Half the speed in 30 cm leaves only a quarter of the energy, which lasts 10 cm — not another 30 cm.

Using each car's own stopping distance

Whether two cars collide depends on their RELATIVE motion: the gap must cover (v_A − v_B)²/2a, not v_A²/2a.

Drill 3 more on braking, penetration and avoiding collision

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Lifts, Blocks in Contact and Pulleys
Second law
$\sum \vec{F} = m\vec{a}, \qquad N_{\text{lift}} = m(g \pm a)$
Force, Work, Power and Circular Dynamics
Work and power
$W = \vec{F}\cdot\vec{r}, \qquad P = \vec{F}\cdot\vec{v}, \qquad F_c = \frac{mv^2}{r}$
Braking, Penetration and Avoiding Collision
Uniform retardation
$v^2 = u^2 - 2as, \qquad s_{\text{safe}} = \frac{(v_A - v_B)^2}{2a}$

Watch out for (6)

Subtracting a for a lift accelerating up
→ Lifts, Blocks in Contact and Pulleys
Using the applied force for the far block
→ Lifts, Blocks in Contact and Pulleys
Taking tension at the bottom as mg
→ Force, Work, Power and Circular Dynamics
Using average power for power at an instant
→ Force, Work, Power and Circular Dynamics
Assuming speed falls linearly with distance
→ Braking, Penetration and Avoiding Collision
Using each car's own stopping distance
→ Braking, Penetration and Avoiding Collision

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