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  4. A wooden block of mass 8 kg slides down an inclined plane of inclination 30° to the horizontal with constant acceleration 0.4 m/s2. The force of friction between the block and inclined plane is (g = 10 m/s2)

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Q. A wooden block of mass $8 \,kg$ slides down an inclined plane of inclination $30^{\circ}$ to the horizontal with constant acceleration $0.4 \,m/s^2$. The force of friction between the block and inclined plane is $(g = 10 \,m/s^2)$

MHT CETMHT CET 2014

Solution:

Here, $m g \sin \theta-f=m a$
image
$m g \sin \theta-m a=f$
$8 \times 10 \sin \left(30^{\circ}\right)-8 \times 0.4=f$
$40-3.2=f \Rightarrow f=36.8 \,N$