Q.A body of unit mass (1kg) moves in a plane, described by two velocity–time graphs. The x-velocity graph is triangular: vx rises linearly from 0 at t=0 to 2m s−1 at t=1s, then falls linearly back to 0 at t=2s, and remains 0 for t>2s. The y-velocity graph rises linearly from 0 at t=0 to 1m s−1 at t=1s, and then stays constant at 1m s−1 for t>1s. Find the force acting on the body as a function of time.
Newton's Second Law: The Law That Connects Force and Motion
Imagine you're pushing a shopping cart. If you push gently, it moves slowly. Push harder, and it speeds up faster. Now imagine the cart is full of groceries — even with the same push, it accelerates much more slowly than an empty cart. This everyday experience is exactly what Newton's Second Law captures.
The Intuition First
Two things matter when you push something:
How hard you push — the force you apply.
How heavy the object is — its mass.
The harder you push, the more the object speeds up. The heavier the object, the less it speeds up for the same push. So acceleration depends on both force and mass — and in opposite ways.
Note
"Acceleration" here means any change in velocity — speeding up, slowing down, or changing direction. It's not just "going faster."
The Precise Statement
Newton's Second Law says:
The acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. The acceleration is in the same direction as the net force.
In one equation:
a=mFnet
Or more commonly:
Fnet=ma
Where:
Fnet is the net force (the vector sum of all forces acting on the object) — measured in newtons (N)
m is the mass of the object — measured in kilograms (kg)
a is the acceleration — measured in metres per second squared (m/s2)
Fnet=ma
What This Really Means
Force causes acceleration, not velocity. A constant net force produces constant acceleration — meaning the velocity keeps changing at a steady rate. If you stop pushing, the net force becomes zero, and acceleration becomes zero (the object continues at constant velocity — that's Newton's First Law).
Mass is a measure of inertia. The more mass an object has, the harder it is to change its motion. A truck needs a much larger force than a bicycle to achieve the same acceleration.
Direction matters. Force and acceleration are vectors — they point the same way. If you push north, the acceleration is north. If multiple forces act, you must add them as vectors to find the net force.
A Simple Example
A 2 kg block is pushed with a net force of 10 N to the right.
a=mFnet=2kg10N=5m/s2
The block accelerates at 5m/s2 to the right. Every second, its velocity increases by 5 m/s in that direction. …
Acceleration is the slope of a velocity–time graph, and with m=1kg the force numerically equals the acceleration. Reading the slopes piecewise gives a force that changes in three time intervals.
Concept
F=ma with m=1kg, and ax=dtdvx, ay=dtdvy are the slopes of the two graphs.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KCET 2024Set D-21 markMCQ
Q.A block of certain mass is placed on a rough inclined plane. The angle between the plane and the horizontal is 30∘. The coefficients of static and kinetic frictions between the block and the inclined plane are 0.6 and 0.5 respectively. Then the magnitude of the acceleration of the block is [Take g=10ms−2]
(A) 2ms−2
(B) zero
(C) 0.196ms−2
(D) 0.67ms−2
›Reveal solutionSolution
Compare the driving force mgsinθ with the maximum available static friction μsmgcosθ — if friction wins, nothing moves and a=0; the kinetic coefficient is then a red herring.
1. Forces along the incline
Resolve the weight of the block (m unknown, but it will cancel):
Driving component down the slope: mgsinθ
Normal reaction: N=mgcosθ
Maximum static friction available: fs,max=μsN=μsmgcosθ
2. The sliding condition
Motion begins only when the driving force exceeds the maximum static friction:
mgsinθ>μsmgcosθ⟺tanθ>μs
This is why the angle of repose is θr=tan−1μs — the mass drops out entirely.
3. Substitute the given data
tanθ=tan30∘=31=0.577
μs=0.6
tan30∘=0.577<0.6=μs
The angle of repose here is θr=tan−1(0.6)=30.96∘, and the plane is tilted only 30∘ — less than the angle of repose.
Q.A charged particle is moving in an electric field of 3×10−10Vm−1 with mobility 2.5×10−6m2/Vs, its drift velocity is
(A) 2.5×10−4m/s
(B) 1.2×10−4m/s
(C) 7.5×10−4m/s
(D) 8.33×10−4m/s
›Reveal solutionSolution
Drift velocity is just mobility times field, vd=μE — a one-line application of the definition of mobility.
Step 1 — The concept.
Inside a conductor the free carriers do not accelerate forever; they are scattered by the lattice every τ seconds, so they acquire only a small steady drift velocity superimposed on their random thermal motion:
vd=mqEτ.
The carrier property mqτ is bundled into a single measured constant called the mobility:
μ=E∣vd∣=mqτ⟹vd=μE
Its SI unit, m2V−1s−1, is exactly (ms−1)/(Vm−1) — velocity per unit field — which tells you at a glance that the two just multiply.
Step 2 — Substitute.
vd=μE=(2.5×10−6m2V−1s−1)×E.
The numerical coefficients multiply as 2.5×3=7.5, so the drift velocity must be of the form 7.5×10nms−1. Among the four options only one has the coefficient 7.5: