Block M = 7:50 kg is initially moving up the incline and is increasing speed wit
ID: 2191348 • Letter: B
Question
Block M = 7:50 kg is initially moving up the incline and is increasing speed with a = 3.68 m/s^2. The applied force F is horizontal. The coecients of friction between the block and incline are s = 0.443 and k = 0.312. The angle of the incline is 25.0 degrees.
(a) What is the force F (N)?
(b) What is the normal force N (in N) between the block and incline?
(c) What is the force of friction (N) on the block?
Explanation / Answer
a)Horizontal direction: F(cos(Radians(25))+(0.312*sin(Radians(25))))-(7.5*3.68)-(7.5*9.81*sin(radians(25)))-(7.5*0.312*9.81*cos(radians(25))) = 0 =>F = 76.57628655 N b)N = (76.57628655*sin(Radians(25)))+(7.5*9.81*cos(radians(25))) = 99.04413254 N c)Force of friction = N*0.312 = 30.90176935 N