Question

⦁ A 46 kg box is pulled on a floor using a string attached to the box and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.50, what minimum force magnitude is required from the rope to start the box moving? (b) If µk =0.35 (coefficient of kinetic friction), what is the magnitude of the initial acceleration of the box?

Answer 1

Answer:

Explanation:

Resolve in vertical and horizontal directions.

m*g - T Sin(15) = Fv (Fv is vertical force) . . . . . EQ1

T Cos(15) = Fv*mu (mu coeff of friction)

so Fv = T*cos(15)/mu . . . . . . . . . . . . . . . . EQ2

Substitute Fv from EQ2 in EQ1

m*g - T*sin(15) = T*cos(15) / mu

Insert given values:

46*9,81 - T*sin(15) = T*cos(15) / 0.5

451.26 - 0.2588*T = 1.93185*T

T = 451.26 / (1.93185 + 0.2588) = 205.99 N or ~ 206 N

hoping thats helpful!!!!!!

Answer 2

Hello there !!

1 km =1000 m total distance= 100 m + 1000, = 1100 m speed = 60 km /hr= 60 x 5/18 = 16.66 m/s time = distance / speed