A 10 kg block rests on a rough horizontal surface (μ=0.1). A force 100 N is applied at 37° to the horizontal as in the source. Find frictional force (N), with g=10 m/s².
Physics · Mixed Chapter
JEE Main Practice

Question

A 10 kg block rests on a rough horizontal surface (μ=0.1). A force 100 N is applied at 37° to the horizontal as in the source. Find frictional force (N), with g=10 m/s².

G1 Physics · Accepted Answer

Step 1: With force inclined 37° upward, its vertical component is Fsin37°=60 N. Step 2: Normal reaction N=mg−Fsin37°=100−60=40 N. Step 3: Limiting/sliding friction f=μN=0.1×40=4 N. Step 4: If the force points downward instead, the answer changes. The image/arrow direction is absent in this PDF section, so upward inclination is inferred from key 4.