Question

k weighing 10.0 newtons is on a ramp inclined 3)A block at 30.0° to the horizontal. A 3.0-newton force of ramp at constant velocity with force F, which is parallel to the ramp, as shown in the diagram below ockfriction, F, acts on the block as it is pulledu 30 10.oN 30.0 (Vectors not drawn to scale) What is the tude of force F? A) 7.0N B) 8.0N 10N D) 13 N 4. The dia ow shows a 10.0-kilogram mass held at rest on a frictionless 30.0° incline by force F 10.0 kg F 30.0° What is the approximate pegnitude ofnforce F? 4)9.81 N B) 49.1 C) 85.0 N In the diagram below, the weight of a box on a plane inclined at 30. is represented by the vector W ) 98.1 N What is the magnitude of he weight (W) that acts parallel to the inclihe? A) W component of the C) 0.87 W 1.5 W

Answer 1

similar to the question 5

Form F.B.D. - F - W' - F_+ = ma Since velocity is const therefore a = 0 therefore F = 10 sin 30 degrees + 3 = 5 + 3 = 8 N (4) theta = 30 degrees, W = mg = 98.2 N therefore F = W sin theta = 98.2 sin 30 degrees rightdoublearrow F = 49.1 N

Magnitude of component of weight parallel to inclination is w Sin30 = 0.5W

Option b.