Question

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.

Answer 1

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