
For hanging block,
40−T=4a⋯(1)
Now, maximum value of friction force will be,
fmax=μN=0.02×40×10=8N
For block on table, T−8=40a⋯(2)
From equation(1) and equation (2), we get32=44a⇒a=4432=118ms−2
A block of mass 40kg slides over a surface, when a mass of 4kg is suspended through an inextensible massless string passing over frictionless pulley as shown below.
The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is: (Given g=10ms−2.)

Held on 29 Jun 2022 · Verified 6 Jul 2026.
118ms−2
1ms−2
51ms−2
54ms−2
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$. 
A particle of mass 2 kg is projected vertically upward with a speed of 30 m/s. The maximum height reached by the particle is (g = 10 m/s²):
Two projectiles are projected with the same initial velocities at the $15^\circ$ and $30^\circ$ with respect to the horizontal. The ratio of their ranges is $1:x$. The value of $x$ is:
In an experiment the values of two spring constants were measured as $k_{1}=(10 \pm 0.2) \mathrm{N} / \mathrm{m}$ and $k_{2}=(20 \pm 0.3) \mathrm{N} / \mathrm{m}$. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :
The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _____. ($\pi = 22/7$)
Work through every JEE Main Mechanics PYQ, year by year.