A 70.0 kg weight sits at the top of a ramp that makes an angle of 40o with the horizontal. As the weight slides down the ramp, it experiences a force of friction of 75 N. What is the acceleration of the weight down the ramp?
From the Newton's second law:
"ma=mgsin\\alpha-F_{fr},"
where m - the mass of a weight;
"\\alpha" - angle, that the ramp makes with the horizontal;
"F_{fr} -" a force of friction.
Hence,
"a=\\frac{mgsin\\alpha-F_{fr}}{m}=\\frac{70\\cdot 9.81\\cdot sin40\\degree-75}{70}=5.23\\space m\/s^2."
Answer: 5.23 "m\/s^2."
Comments
Leave a comment