A box is moving with an initial velocity v = 25 m/s. If the coefficient of kinetic friction between the floor and the box is μ = 0.3 how far will the box travel before coming to rest?
Please explain or provide the equations you used if possible! Thanks!
Please explain or provide the equations you used if possible! Thanks!
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let mass be m
friction = mg * 0.3
= m * acc
acc = 0.3 *9.81 = 2.943m/s2
u=25; v=0
v2 = u2 + 2as
0 = 25^2 - 2* 2.943 *s
s = 106.18 m
answer
note: acc is actually deceleration as a result of friction; thus negative value
friction = mg * 0.3
= m * acc
acc = 0.3 *9.81 = 2.943m/s2
u=25; v=0
v2 = u2 + 2as
0 = 25^2 - 2* 2.943 *s
s = 106.18 m
answer
note: acc is actually deceleration as a result of friction; thus negative value
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You'd need to know the mass or weight of the box in order to answer this question.