A 24.0-kg box slides down an incline that is 17.0 m long and tilted with an angl

Question

A 24.0-kg box slides down an incline that is 17.0 m long and tilted
with an angle of 35.0? relative to the horizontal. Assume that the
box is released from rest at the highest point of the incline and the
coefficient of kinetic friction between the box and the incline is 0.200.
Use g = 9.80 m/s2.
(9) What is the magnitude of the normal force on the box?
(A) 157 N; (B) 166 N; (C) 175 N; (D) 184 N; (E) 193 N.
(10) What is the magnitude of the kinetic friction on the box?
(A) 36.8 N; (B) 35.0 N; (C) 33.2 N; (D) 38.5 N; (E) 31.4 N.
(11) What is the magnitude of the acceleration of the box?
(A) 3.04 m/s2; (B) 3.53 m/s2; (C) 4.02 m/s2; (D) 4.51 m/s2;
(E) 5.00 m/s2.
(12) How long does it take for the box to slide through the incline?
(A) 2.91 s; (B) 3.22 s; (C) 3.53 s; (D) 3.84 s; (E) 4.15 s.

Explanation / Answer

9) E)193 N

N = m*g*cos(35) = 24*9.8*cos(35) = 193 N

10) D) 38.5 N

F_k = mue_k*N = 0.2*193 = 38.5 N

11) C) 4.02 m/s^2

a = g*sin(35) - mue_k*g*cos(35)

= 9.8*sin(35) - 0.2*9.8*cos(35)

= 4.02 m/s^2

12) A)2.91 s

s = 0.5*a*t^2

t = sqrt(2*s/a)

= sqrt(2*17/4.02)

= 2.91 s

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