A solid block of mass 2 {kg} is resting inside a cube as shown in the figure. The cube is moving with the velocity \overrightarrow{v}=5 t \hat{i}+2 \hat{j} {m} \/ {s}(t is time in second). The block is at rest with respect to the cube and coefficient of friction between the surfaces of cube and the block is 0.6. Then (g=10 {m} \/ {s}^{2})

force of friction acting on the block is 10 {N}.

force of friction acting on the block is 4 {N}.

the total force exerted by the block on the cube is 14 {N}.

the total force exerted by the block on the cube is 10 \sqrt{5} {N}.

JEE Advanced 2020 Concept Recapitulation Test 3 Paper 1

Question : 6 of 54

Marks: +1, -0

A solid block of mass 2kg is resting inside a cube as shown in the figure. The cube is moving with the velocity

→

=5t

∧

m∕s(t is time in second). The block is at rest with respect to the cube and coefficient of friction between the surfaces of cube and the block is 0.6. Then (g=10m∕s2)

force of friction acting on the block is 10N.
force of friction acting on the block is 4N.
the total force exerted by the block on the cube is 14N.
the total force exerted by the block on the cube is 10√5N.

Validate

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