A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 171 n at an angle of 30.7 above the horizontal. The box has a mass of 38.9 kg, and k between the box and the floor is 0.19.
Answers
Answer:
Mass of the box an books = 43.7 kg
Coefficient of kinetic friction = 0.222
Force = 207 N
Angle = 46.7°
(A). we draw the free body diagram
(B). We need to calculate the normal force
Using formula of normal force
∑y=0
N+f\sin\theta=mgN+fsinθ=mg
N=mg-f\sin\thetaN=mg−fsinθ
Put the value into the formula
N=43.7\times9.8-207\sin46.7N=43.7×9.8−207sin46.7
N=277.61\ NN=277.61 N
The normal force that the floor exerts on the box is 277.61 N.
(c). We need to calculate the acceleration of the box
Using formula of acceleration
\Sigma F_{x}=maΣF
x
=ma
F\cos\theta-f_{k}=maFcosθ−f
k
=ma
a=\dfrac{F\cos\theta-\mu_{k}N}{m}a=
m
Fcosθ−μ
k
N
Put the value into the formula
a=\dfrac{207\cos46.7-0.222\times277.61}{43.7}a=
43.7
207cos46.7−0.222×277.61
a=1.84\ m/s^2a=1.84 m/s
2
The acceleration of the box is 1.84 m/s².
Hence, This is the required solution.
Explanation:
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