A 12.0-kg block is pushed across a rough horizontal surface by a force that is angled 30.0◦ below the horizontal. The magnitude of the force is 75.0 N and the acceleration of the block as it is pushed is 3.20 m/s2. What is the magnitude of the contact force exerted on the block by the surface?

157.36 N

Explanation:

Contact force is the force which is created due to contact and it is applied on the contact point . The force applied by body on the surface is its weight .

If R be the reaction force of the ground

R = mg + F son30

= 12 x 9.8 + 75 sin 30

= 117.6 + 37.5

= 155.10 N .

friction force = f

Net force in forward direction = F cos 30 - f  = ma

75cos 30 - f = 12 x 3.2

f = 65 - 38.4

= 26.6 N  

Total force on the surface =√( f² + R² )

√ (26.6² + 155.1²)

= √707.56 + 24056²

=√ 24763.57

= 157.36 N.

contact force = 157.36 N .

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