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14 June, 10:26

A block of mass 9.5kg rests on a slope an angle of 23.0∘ relative to the horizontal. What is the size of the contact force normal to the slope, in Newtons?

Use a gravitational acceleration value of g = 9.8m/s^2.

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  1. 14 June, 11:29
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    The Normal Force = M x G x Cos (theta)

    = 9.5 Kg x 9.8 m/s^2 x cos 23

    = 9.5 Kg x 9.8 m/s^2 x 0.9205

    Converting Kg to Newton,

    1 Kg = 9.81 N

    = 9.5 Kg x 9.81 N x 9.8 m/s^2 x 0.9205

    = 840.702 N
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