Question
A block of mass 21.0kg kg slides down a frictionless surface inclined at 33 ∘ ^\circ.To ensure that the block does not accelerate, what is the magnitude of the smallest force that you must exert on it?
Answers
Henry
Wb = m*g = 21kg * 9.8N/kg = 205.8 N. =
Wt. of the block.
Fp = 205.8*sin33 = 112.1 N. = Force
parallel to the incline.
Fe-Fp = m*a = m*0 = 0
Fe - 112.1 = 0
Fe = 112.1 N. = Force exerted.
Wt. of the block.
Fp = 205.8*sin33 = 112.1 N. = Force
parallel to the incline.
Fe-Fp = m*a = m*0 = 0
Fe - 112.1 = 0
Fe = 112.1 N. = Force exerted.